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Understanding the Hardness of ProvingFormulas in Propositional Logic
Jakob Nordstrom
KTH Royal Institute of TechnologyStockholm, Sweden
Proof complexity workshopBanff International Research Station, Canada
October 2–7, 2011
Joint work with Eli Ben-Sasson
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 1 / 37
A Fundamental Theoretical Problem. . .
Problem
Given a propositional logic formula F , is it true no matter how weassign values to its variables?
tautology: Fundamental problem in Theoretical ComputerScience since Cook’s NP-completeness paper (1971)
(And significance realized much earlier — cf. Godel’s letter 1956)
These days recognized as one of the main challenges for all ofmathematics — one of the million dollar “Millennium Problems”
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 2 / 37
. . . with Huge Practical Implications
All known algorithms run in exponential time in worst case
But enormous progress on applied computer programs last10-15 years
These so-called SAT solvers are routinely deployed to solvelarge-scale real-world problems with millions of variables
Used in e.g. hardware verification, software testing, softwarepackage management, artificial intelligence, cryptography,bioinformatics, and more
But we also know small example formulas with only hundredsof variables that trip up even state-of-the-art SAT solvers
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 3 / 37
What Makes Formulas Hard or Easy?
Best known algorithms based on simple DPLL method from1960s (although with many clever optimizations)
Corresponds to search algorithm for resolution proof system
How can SAT solvers be so good in practice? And whatexplains whether a particular formula is tractable or too hard?
Key bottlenecks for SAT solvers: time and memory
What are the connections between these resources?Are they correlated? Are there trade-offs?
This talk: What can proof complexity say about this? (Forresolution and more general k-DNF resolution proof systems)
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 4 / 37
Outline
1 Resolution-Based Proof SystemsBasicsSome Previous WorkOur Results
2 Outline of ProofsPebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
3 Open ProblemsSpace in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 5 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Some Notation and Terminology
Literal a: variable x or its negation x
Clause C = a1 ∨ · · · ∨ ak: disjunction of literals
Term T = a1 ∧ · · · ∧ ak: conjunction of literals
CNF formula F = C1 ∧ · · · ∧ Cm: conjunction of clausesk-CNF formula: CNF formula with clauses of size ≤ k
DNF formula D = T1 ∨ · · · ∨ Tm: disjunction of termsk-DNF formula: DNF formula with terms of size ≤ k
All CNF formulas assumed to have clauses of size O(1)throughout this talk
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 6 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
k-DNF Resolution
Prove that given CNF formula is unsatisfiable
Proof operates with k-DNF formulas (standard resolutioncorresponds to 1-DNF formulas, i.e., disjunctive clauses)
Proof is “presented on blackboard”
Derivation steps:
Write down clauses of CNF formula being refuted(axiom clauses)Infer new k-DNF formulasErase formulas that are not currently needed (to save space onblackboard)
Proof ends when contradictory empty clause 0 derived
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 7 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
xWrite down axiom 1: x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
x
y ∨ z
Write down axiom 1: xWrite down axiom 3: y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
x
y ∨ z
Write down axiom 1: xWrite down axiom 3: y ∨ zCombine x and y ∨ z
to get (x ∧ y) ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
x
y ∨ z
(x ∧ y) ∨ z
Write down axiom 1: xWrite down axiom 3: y ∨ zCombine x and y ∨ z
to get (x ∧ y) ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
x
y ∨ z
(x ∧ y) ∨ z
Write down axiom 1: xWrite down axiom 3: y ∨ zCombine x and y ∨ z
to get (x ∧ y) ∨ zErase the line x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
y ∨ z
(x ∧ y) ∨ z
Write down axiom 1: xWrite down axiom 3: y ∨ zCombine x and y ∨ z
to get (x ∧ y) ∨ zErase the line x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
y ∨ z
(x ∧ y) ∨ z
Write down axiom 3: y ∨ zCombine x and y ∨ z
to get (x ∧ y) ∨ zErase the line xErase the line y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
(x ∧ y) ∨ zWrite down axiom 3: y ∨ zCombine x and y ∨ z
to get (x ∧ y) ∨ zErase the line xErase the line y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
(x ∧ y) ∨ z
x ∨ y
Combine x and y ∨ zto get (x ∧ y) ∨ z
Erase the line xErase the line y ∨ zWrite down axiom 2: x ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
(x ∧ y) ∨ z
x ∨ y
Erase the line xErase the line y ∨ zWrite down axiom 2: x ∨ yInfer z from
x ∨ y and (x ∧ y) ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
(x ∧ y) ∨ z
x ∨ y
z
Erase the line xErase the line y ∨ zWrite down axiom 2: x ∨ yInfer z from
x ∨ y and (x ∧ y) ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
(x ∧ y) ∨ z
x ∨ y
z
Erase the line y ∨ zWrite down axiom 2: x ∨ yInfer z from
x ∨ y and (x ∧ y) ∨ zErase the line (x ∧ y) ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
x ∨ y
z
Erase the line y ∨ zWrite down axiom 2: x ∨ yInfer z from
x ∨ y and (x ∧ y) ∨ zErase the line (x ∧ y) ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
x ∨ y
z
Write down axiom 2: x ∨ yInfer z from
x ∨ y and (x ∧ y) ∨ zErase the line (x ∧ y) ∨ zErase the line x ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
zWrite down axiom 2: x ∨ yInfer z from
x ∨ y and (x ∧ y) ∨ zErase the line (x ∧ y) ∨ zErase the line x ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
z
z
Infer z fromx ∨ y and (x ∧ y) ∨ z
Erase the line (x ∧ y) ∨ zErase the line x ∨ yWrite down axiom 4: z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
z
z
Erase the line (x ∧ y) ∨ zErase the line x ∨ yWrite down axiom 4: zInfer 0 from
z and z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 2-DNF Resolution Refutation
Can write down axioms,infer new formulas, anderase used formulas
1. x2. x ∨ y3. y ∨ z4. z
Rules:
Infer new formulas only fromformulas currently on board
Only k-DNF formulas canappear on board (for k = 2)
Details about derivation ruleswon’t matter for us
z
z
0
Erase the line (x ∧ y) ∨ zErase the line x ∨ yWrite down axiom 4: zInfer 0 from
z and z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 8 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Complexity Measures of Interest: Length and Space
Length: Lower bound on time for proof search algorithm
Space: Lower bound on memory for proof search algorithm
Length# formulas written on blackboard counted with repetitions
SpaceSomewhat less straightforward — several ways of measuring
1.
x
1
2.
y
2
∨ z
3
3.
(x
4
∧ y)
5
∨ z
6
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 9 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Complexity Measures of Interest: Length and Space
Length: Lower bound on time for proof search algorithm
Space: Lower bound on memory for proof search algorithm
Length# formulas written on blackboard counted with repetitions
SpaceSomewhat less straightforward — several ways of measuring
1.
x
1
2.
y
2
∨ z
3
3.
(x
4
∧ y)
5
∨ z
6
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 9 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Complexity Measures of Interest: Length and Space
Length: Lower bound on time for proof search algorithm
Space: Lower bound on memory for proof search algorithm
Length# formulas written on blackboard counted with repetitions
SpaceSomewhat less straightforward — several ways of measuring
1.
x
1
2.
y
2
∨ z
3
3.
(x
4
∧ y)
5
∨ z
6
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 9 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Complexity Measures of Interest: Length and Space
Length: Lower bound on time for proof search algorithm
Space: Lower bound on memory for proof search algorithm
Length# formulas written on blackboard counted with repetitions
SpaceSomewhat less straightforward — several ways of measuring
1. x
1
2. y
2
∨ z
3
3. (x
4
∧ y)
5
∨ z
6
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 9 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Complexity Measures of Interest: Length and Space
Length: Lower bound on time for proof search algorithm
Space: Lower bound on memory for proof search algorithm
Length# formulas written on blackboard counted with repetitions
SpaceSomewhat less straightforward — several ways of measuring
1.
x1
2.
y2∨ z3
3.
(x4∧ y)5∨ z6
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 9 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Complexity Measures of Interest: Length and Space
Length: Lower bound on time for proof search algorithm
Space: Lower bound on memory for proof search algorithm
Length# formulas written on blackboard counted with repetitions
SpaceSomewhat less straightforward — several ways of measuring
1.
x1
2.
y2∨ z3
3.
(x
4
∧ y)
5
∨ z
6
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 9 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Complexity Measures of Interest: Length and Space
Length: Lower bound on time for proof search algorithm
Space: Lower bound on memory for proof search algorithm
Length# formulas written on blackboard counted with repetitions
SpaceSomewhat less straightforward — several ways of measuring
1.
x
1
2.
y
2
∨ z
3
3.
(x
4
∧ y)
5
∨ z
6
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 9 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Length and Space Bounds for Resolution
Let n = size of formula
Length: at most 2n
Lower bound exp(Ω(n)) [Urquhart ’87, Chvatal & Szemeredi ’88]
Formula space (a.k.a. clause space): at most nLower bound Ω(n) [Toran ’99, Alekhnovich et al. ’00]
Total space: at most n2
No better lower bound than Ω(n)!?
[Sidenote: space bounds hold even for “magic algorithms” alwaysmaking optimal choices — so might be much stronger in practice]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 10 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Comparing Length and Space
Some “rescaling” needed to get meaningful comparisons of lengthand space
Length exponential in formula size in worst case
Formula space at most linear
So natural to compare space to logarithm of length
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 11 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Length-Space Correlation for Resolution?
∃ constant space refutation ⇒ ∃ polynomial length refutation[Atserias & Dalmau ’03]
For tree-like resolution: any polynomial length refutation can becarried out in logarithmic space [Esteban & Toran ’99]
So essentially no trade-offs for tree-like resolution
Does short length imply small space for general resolution?Has been open — even no consensus on likely “right answer”
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 12 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Length-Space Correlation for Resolution?
∃ constant space refutation ⇒ ∃ polynomial length refutation[Atserias & Dalmau ’03]
For tree-like resolution: any polynomial length refutation can becarried out in logarithmic space [Esteban & Toran ’99]
So essentially no trade-offs for tree-like resolution
Does short length imply small space for general resolution?Has been open — even no consensus on likely “right answer”
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 12 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Length-Space Correlation for Resolution?
∃ constant space refutation ⇒ ∃ polynomial length refutation[Atserias & Dalmau ’03]
For tree-like resolution: any polynomial length refutation can becarried out in logarithmic space [Esteban & Toran ’99]
So essentially no trade-offs for tree-like resolution
Does short length imply small space for general resolution?Has been open — even no consensus on likely “right answer”
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 12 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Length-Space Trade-offs for Resolution?
Nothing known about length-space trade-offs for resolutionrefutations in the general, unrestricted proof system
(Some trade-off results in restricted settings in [Ben-Sasson ’02,Nordstrom ’07])
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 13 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Previous Work on k-DNF Resolution (k ≥ 2)
Length: lower bound exp(Ω
(n1−o(1)
))[Segerlind et al. ’04,
Alekhnovich ’05]
Formula space: lower bound Ω(n)
[Esteban et al. ’02]
(Suppressing dependencies on k)
(k+1)-DNF resolution exponentially stronger thank-DNF resolution w.r.t. length [Segerlind et al. ’04]
No hierarchy known w.r.t. spaceExcept for tree-like k-DNF resolution [Esteban et al. ’02](But tree-like k-DNF weaker than standard resolution)
No trade-off results known
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 14 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Previous Work on k-DNF Resolution (k ≥ 2)
Length: lower bound exp(Ω
(n1−o(1)
))[Segerlind et al. ’04,
Alekhnovich ’05]
Formula space: lower bound Ω(n)
[Esteban et al. ’02]
(Suppressing dependencies on k)
(k+1)-DNF resolution exponentially stronger thank-DNF resolution w.r.t. length [Segerlind et al. ’04]
No hierarchy known w.r.t. spaceExcept for tree-like k-DNF resolution [Esteban et al. ’02](But tree-like k-DNF weaker than standard resolution)
No trade-off results known
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 14 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Previous Work on k-DNF Resolution (k ≥ 2)
Length: lower bound exp(Ω
(n1−o(1)
))[Segerlind et al. ’04,
Alekhnovich ’05]
Formula space: lower bound Ω(n)
[Esteban et al. ’02]
(Suppressing dependencies on k)
(k+1)-DNF resolution exponentially stronger thank-DNF resolution w.r.t. length [Segerlind et al. ’04]
No hierarchy known w.r.t. spaceExcept for tree-like k-DNF resolution [Esteban et al. ’02](But tree-like k-DNF weaker than standard resolution)
No trade-off results known
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 14 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Our results 1: An Optimal Length-Space Separation
Length and space in resolution are “completely uncorrelated”
Theorem (Ben-Sasson & Nordstrom, FOCS ’08)
There are k-CNF formula families of size O(n) with
refutation length O(n) requiring
formula space Ω(n/ log n).
Optimal separation of length and space — given length n, alwayspossible to achieve space O(n/ log n)
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 15 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Our Results 2: Length-Space Trade-offs
We prove collection of length-space trade-offs
Results hold for
resolution (essentially tight analysis)
k-DNF resolution, k ≥ 2 (with slightly worse parameters)
Different trade-offs covering (almost) whole range of space fromconstant to linear
Simple, explicit formulas
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 16 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 1: Robust Trade-offs for Small Space
Theorem (Ben-Sasson & Nordstrom, ICS ’11)
For any ω(1) function and any fixed K there exist explicitCNF formulas of size O(n)
refutable in resolution in total space ω(1)
refutable in resolution in length O(n) and total space ≈ 3√
n
any resolution refutation in formula space 3√
n requiressuperpolynomial length
any k-DNF resolution refutation, k ≤ K, in formula space n1/3(k+1) requires superpolynomial length
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 17 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 1: Robust Trade-offs for Small Space
Theorem (Ben-Sasson & Nordstrom, ICS ’11)
For any ω(1) function and any fixed K there exist explicitCNF formulas of size O(n)
refutable in resolution in total space ω(1)
refutable in resolution in length O(n) and total space ≈ 3√
n
any resolution refutation in formula space 3√
n requiressuperpolynomial length
any k-DNF resolution refutation, k ≤ K, in formula space n1/3(k+1) requires superpolynomial length
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 17 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 1: Robust Trade-offs for Small Space
Theorem (Ben-Sasson & Nordstrom, ICS ’11)
For any ω(1) function and any fixed K there exist explicitCNF formulas of size O(n)
refutable in resolution in total space ω(1)
refutable in resolution in length O(n) and total space ≈ 3√
n
any resolution refutation in formula space 3√
n requiressuperpolynomial length
any k-DNF resolution refutation, k ≤ K, in formula space n1/3(k+1) requires superpolynomial length
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 17 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 1: Robust Trade-offs for Small Space
Theorem (Ben-Sasson & Nordstrom, ICS ’11)
For any ω(1) function and any fixed K there exist explicitCNF formulas of size O(n)
refutable in resolution in total space ω(1)
refutable in resolution in length O(n) and total space ≈ 3√
n
any resolution refutation in formula space 3√
n requiressuperpolynomial length
any k-DNF resolution refutation, k ≤ K, in formula space n1/3(k+1) requires superpolynomial length
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 17 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Example 1: Robust Trade-offs for Small Space
Theorem (Ben-Sasson & Nordstrom, ICS ’11)
For any ω(1) function and any fixed K there exist explicitCNF formulas of size O(n)
refutable in resolution in total space ω(1)
refutable in resolution in length O(n) and total space ≈ 3√
n
any resolution refutation in formula space 3√
n requiressuperpolynomial length
any k-DNF resolution refutation, k ≤ K, in formula space n1/3(k+1) requires superpolynomial length
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 17 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Some Quick Technical Remarks
Upper bounds hold for
total space (# literals) — larger measure
standard syntactic rules
Lower bounds hold for
formula space (# lines) — smaller measure
semantic rules — exponentially stronger than syntactic
Space definition reminder
x
y ∨ z
(x ∧ y) ∨ z
Formula space: 3Total space: 6Variable space: 3
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 18 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
BasicsSome Previous WorkOur Results
Our Results 3: Space Hierarchy for k-DNF Resolution
We also separate k-DNF resolution from (k+1)-DNF resolutionw.r.t. formula space
Theorem (Ben-Sasson & Nordstrom, ICS ’11)
For any constant k there are explicit CNF formulas of size O(n)
refutable in (k+1)-DNF resolution in formula space O(1) butsuch that
any k-DNF resolution refutation requires formula spaceΩ
(k+1√
n/ log n)
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 19 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Rest of This Talk
Study old combinatorial game from the 70s and 80s
Prove new theorem about amplification of space hardness viavariable substitution
Combine the two
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 20 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
How to Get a Handle on Time-Space Relations?
Want to find formulas that
can be quickly refuted but require large space
have space-efficient refutations requiring much time
Such time-space trade-off questions well-studied forpebble games modelling calculations described by DAGs([Cook & Sethi ’76] and many others)
Time needed for calculation: # pebbling moves
Space needed for calculation: max # pebbles required
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 21 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 0
Current # pebbles 0
Max # pebbles so far 0
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 1
Current # pebbles 1
Max # pebbles so far 1
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 2
Current # pebbles 2
Max # pebbles so far 2
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 3
Current # pebbles 3
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 4
Current # pebbles 2
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 5
Current # pebbles 1
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 6
Current # pebbles 2
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 7
Current # pebbles 3
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 8
Current # pebbles 2
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 8
Current # pebbles 2
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 9
Current # pebbles 3
Max # pebbles so far 3
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 10
Current # pebbles 4
Max # pebbles so far 4
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 11
Current # pebbles 3
Max # pebbles so far 4
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 12
Current # pebbles 2
Max # pebbles so far 4
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
The Black-White Pebble Game
Goal: get single black pebble on sink vertex of G
z
x y
u v w
# moves 13
Current # pebbles 1
Max # pebbles so far 4
1 Can place black pebble on (empty) vertex v if all predecessors(vertices with edges to v) have pebbles on them
2 Can always remove black pebble from vertex
3 Can always place white pebble on (empty) vertex
4 Can remove white pebble if all predecessors have pebbles
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 22 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Pebbling Contradiction
CNF formula encoding pebble game on DAG G
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
sources are true
truth propa-gates upwards
but sink is false
Studied by [Bonet et al. ’98, Raz & McKenzie ’99, Ben-Sasson &Wigderson ’99] and others
Our hope is that pebbling properties of DAG somehow carry overto resolution refutations of pebbling contradictions
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 23 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Pebbling Contradiction
CNF formula encoding pebble game on DAG G
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
sources are true
truth propa-gates upwards
but sink is false
Studied by [Bonet et al. ’98, Raz & McKenzie ’99, Ben-Sasson &Wigderson ’99] and others
Our hope is that pebbling properties of DAG somehow carry overto resolution refutations of pebbling contradictions
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 23 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Interpreting Refutations as Black-White Pebblings
Black-white pebbling models non-deterministic computation
black pebbles ⇔ computed results
white pebbles ⇔ guesses needing to be verified
“Know z assuming v, w”
Corresponds to (v ∧ w) → z, i.e.,blackboard clause v ∨ w ∨ z
So translate clauses to pebbles by:unnegated variable⇒ black pebblenegated variable⇒white pebble
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 24 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Interpreting Refutations as Black-White Pebblings
Black-white pebbling models non-deterministic computation
black pebbles ⇔ computed results
white pebbles ⇔ guesses needing to be verified
z
x y
u v w
“Know z assuming v, w”
Corresponds to (v ∧ w) → z, i.e.,blackboard clause v ∨ w ∨ z
So translate clauses to pebbles by:unnegated variable⇒ black pebblenegated variable⇒white pebble
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 24 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Interpreting Refutations as Black-White Pebblings
Black-white pebbling models non-deterministic computation
black pebbles ⇔ computed results
white pebbles ⇔ guesses needing to be verified
z
x y
u v w
“Know z assuming v, w”
Corresponds to (v ∧ w) → z, i.e.,blackboard clause v ∨ w ∨ z
So translate clauses to pebbles by:unnegated variable⇒ black pebblenegated variable⇒white pebble
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 24 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u Write down axiom 1: u
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u
vWrite down axiom 1: uWrite down axiom 2: v
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u
v
u ∨ v ∨ x
Write down axiom 1: uWrite down axiom 2: vWrite down axiom 4: u ∨ v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u
v
u ∨ v ∨ x
Write down axiom 1: uWrite down axiom 2: vWrite down axiom 4: u ∨ v ∨ xInfer v ∨ x from
u and u ∨ v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u
v
u ∨ v ∨ x
v ∨ x
Write down axiom 1: uWrite down axiom 2: vWrite down axiom 4: u ∨ v ∨ xInfer v ∨ x from
u and u ∨ v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u
v
u ∨ v ∨ x
v ∨ x
Write down axiom 2: vWrite down axiom 4: u ∨ v ∨ xInfer v ∨ x from
u and u ∨ v ∨ xErase the clause u ∨ v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u
v
v ∨ x
Write down axiom 2: vWrite down axiom 4: u ∨ v ∨ xInfer v ∨ x from
u and u ∨ v ∨ xErase the clause u ∨ v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
u
v
v ∨ x
Write down axiom 4: u ∨ v ∨ xInfer v ∨ x from
u and u ∨ v ∨ xErase the clause u ∨ v ∨ xErase the clause u
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v
v ∨ xWrite down axiom 4: u ∨ v ∨ xInfer v ∨ x from
u and u ∨ v ∨ xErase the clause u ∨ v ∨ xErase the clause u
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v
v ∨ xu and u ∨ v ∨ x
Erase the clause u ∨ v ∨ xErase the clause uInfer x from
v and v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v
v ∨ x
x
u and u ∨ v ∨ xErase the clause u ∨ v ∨ xErase the clause uInfer x from
v and v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v
v ∨ x
x
Erase the clause u ∨ v ∨ xErase the clause uInfer x from
v and v ∨ xErase the clause v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v
xErase the clause u ∨ v ∨ xErase the clause uInfer x from
v and v ∨ xErase the clause v ∨ x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v
xErase the clause uInfer x from
v and v ∨ xErase the clause v ∨ xErase the clause v
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
x Erase the clause uInfer x from
v and v ∨ xErase the clause v ∨ xErase the clause v
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
x
x ∨ y ∨ zInfer x from
v and v ∨ xErase the clause v ∨ xErase the clause vWrite down axiom 6: x ∨ y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
x
x ∨ y ∨ zErase the clause v ∨ xErase the clause vWrite down axiom 6: x ∨ y ∨ zInfer y ∨ z from
x and x ∨ y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
x
x ∨ y ∨ z
y ∨ z
Erase the clause v ∨ xErase the clause vWrite down axiom 6: x ∨ y ∨ zInfer y ∨ z from
x and x ∨ y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
x
x ∨ y ∨ z
y ∨ z
Erase the clause vWrite down axiom 6: x ∨ y ∨ zInfer y ∨ z from
x and x ∨ y ∨ zErase the clause x ∨ y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
x
y ∨ zErase the clause vWrite down axiom 6: x ∨ y ∨ zInfer y ∨ z from
x and x ∨ y ∨ zErase the clause x ∨ y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
x
y ∨ zWrite down axiom 6: x ∨ y ∨ zInfer y ∨ z from
x and x ∨ y ∨ zErase the clause x ∨ y ∨ zErase the clause x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
y ∨ z Write down axiom 6: x ∨ y ∨ zInfer y ∨ z from
x and x ∨ y ∨ zErase the clause x ∨ y ∨ zErase the clause x
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
y ∨ z
v ∨ w ∨ yInfer y ∨ z from
x and x ∨ y ∨ zErase the clause x ∨ y ∨ zErase the clause xWrite down axiom 5: v ∨ w ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
y ∨ z
v ∨ w ∨ yErase the clause x ∨ y ∨ zErase the clause xWrite down axiom 5: v ∨ w ∨ yInfer v ∨ w ∨ z from
y ∨ z and v ∨ w ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
y ∨ z
v ∨ w ∨ y
v ∨ w ∨ z
Erase the clause x ∨ y ∨ zErase the clause xWrite down axiom 5: v ∨ w ∨ yInfer v ∨ w ∨ z from
y ∨ z and v ∨ w ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
y ∨ z
v ∨ w ∨ y
v ∨ w ∨ z
Erase the clause xWrite down axiom 5: v ∨ w ∨ yInfer v ∨ w ∨ z from
y ∨ z and v ∨ w ∨ yErase the clause v ∨ w ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
y ∨ z
v ∨ w ∨ zErase the clause xWrite down axiom 5: v ∨ w ∨ yInfer v ∨ w ∨ z from
y ∨ z and v ∨ w ∨ yErase the clause v ∨ w ∨ y
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
y ∨ z
v ∨ w ∨ zWrite down axiom 5: v ∨ w ∨ yInfer v ∨ w ∨ z from
y ∨ z and v ∨ w ∨ yErase the clause v ∨ w ∨ yErase the clause y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z Write down axiom 5: v ∨ w ∨ yInfer v ∨ w ∨ z from
y ∨ z and v ∨ w ∨ yErase the clause v ∨ w ∨ yErase the clause y ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
vInfer v ∨ w ∨ z from
y ∨ z and v ∨ w ∨ yErase the clause v ∨ w ∨ yErase the clause y ∨ zWrite down axiom 2: v
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
v
w
y ∨ z and v ∨ w ∨ yErase the clause v ∨ w ∨ yErase the clause y ∨ zWrite down axiom 2: vWrite down axiom 3: w
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
v
w
z
Erase the clause v ∨ w ∨ yErase the clause y ∨ zWrite down axiom 2: vWrite down axiom 3: wWrite down axiom 7: z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
v
w
z
Write down axiom 2: vWrite down axiom 3: wWrite down axiom 7: zInfer w ∨ z from
v and v ∨ w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
v
w
z
w ∨ z
Write down axiom 2: vWrite down axiom 3: wWrite down axiom 7: zInfer w ∨ z from
v and v ∨ w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
v
w
z
w ∨ z
Write down axiom 3: wWrite down axiom 7: zInfer w ∨ z from
v and v ∨ w ∨ zErase the clause v
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
w
z
w ∨ z
Write down axiom 3: wWrite down axiom 7: zInfer w ∨ z from
v and v ∨ w ∨ zErase the clause v
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
v ∨ w ∨ z
w
z
w ∨ z
Write down axiom 7: zInfer w ∨ z from
v and v ∨ w ∨ zErase the clause vErase the clause v ∨ w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
w
z
w ∨ z
Write down axiom 7: zInfer w ∨ z from
v and v ∨ w ∨ zErase the clause vErase the clause v ∨ w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
w
z
w ∨ z
v and v ∨ w ∨ zErase the clause vErase the clause v ∨ w ∨ zInfer z from
w and w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
w
z
w ∨ z
z
v and v ∨ w ∨ zErase the clause vErase the clause v ∨ w ∨ zInfer z from
w and w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
w
z
w ∨ z
z
Erase the clause vErase the clause v ∨ w ∨ zInfer z from
w and w ∨ zErase the clause w
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
z
w ∨ z
z
Erase the clause vErase the clause v ∨ w ∨ zInfer z from
w and w ∨ zErase the clause w
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
z
w ∨ z
z
Erase the clause v ∨ w ∨ zInfer z from
w and w ∨ zErase the clause wErase the clause w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
z
zErase the clause v ∨ w ∨ zInfer z from
w and w ∨ zErase the clause wErase the clause w ∨ z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
z
zw and w ∨ z
Erase the clause wErase the clause w ∨ zInfer 0 from
z and z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Example of Refutation-Pebbling Correspondence
1. u2. v3. w4. u ∨ v ∨ x5. v ∨ w ∨ y6. x ∨ y ∨ z7. z
z
x y
u v w
z
z
0
w and w ∨ zErase the clause wErase the clause w ∨ zInfer 0 from
z and z
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 25 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Formal Refutation-Pebbling Correspondence
Theorem (Ben-Sasson ’02)
Any refutation translates into black-white pebbling with
# moves ≤ refutation length
# pebbles ≤ variable space
Observation (Ben-Sasson et al. ’00)
Any black-pebbles-only pebbling translates into refutation with
refutation length ≤ # moves
total space ≤ # pebbles
Unfortunately pebbling contradictions are extremely easy w.r.t.formula space! — not what we want
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 26 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Formal Refutation-Pebbling Correspondence
Theorem (Ben-Sasson ’02)
Any refutation translates into black-white pebbling with
# moves ≤ refutation length
# pebbles ≤ variable space
Observation (Ben-Sasson et al. ’00)
Any black-pebbles-only pebbling translates into refutation with
refutation length ≤ # moves
total space ≤ # pebbles
Unfortunately pebbling contradictions are extremely easy w.r.t.formula space! — not what we want
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 26 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Formal Refutation-Pebbling Correspondence
Theorem (Ben-Sasson ’02)
Any refutation translates into black-white pebbling with
# moves ≤ refutation length
# pebbles ≤ variable space
Observation (Ben-Sasson et al. ’00)
Any black-pebbles-only pebbling translates into refutation with
refutation length ≤ # moves
total space ≤ # pebbles
Unfortunately pebbling contradictions are extremely easy w.r.t.formula space! — not what we want
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 26 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Idea: Variable Substitution
Make formula harder by substituting x1 ⊕ x2 for every variable x(also works for other Boolean functions with “right” properties):
x ∨ y
⇓
¬(x1 ⊕ x2) ∨ (y1 ⊕ y2)
⇓
(x1 ∨ x2 ∨ y1 ∨ y2)
∧ (x1 ∨ x2 ∨ y1 ∨ y2)
∧ (x1 ∨ x2 ∨ y1 ∨ y2)
∧ (x1 ∨ x2 ∨ y1 ∨ y2)
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 27 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
x ∨ y
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
x ∨ y
y
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
x ∨ y
y
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
x1 ∨ x2
x1 ∨ x2
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
x ∨ y
y
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
x1 ∨ x2
x1 ∨ x2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
x ∨ y
y
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
x1 ∨ x2
x1 ∨ x2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
y1 ∨ y2
y1 ∨ y2
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
x ∨ y
y
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
x1 ∨ x2
x1 ∨ x2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
y1 ∨ y2
y1 ∨ y2
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Key Technical Result: Substitution Theorem
Let F [⊕] denote formula with XOR x1 ⊕ x2 substituted for x
Obvious approach for F [⊕]: mimic refutation of F
x
x ∨ y
y
For such refutation of F [⊕]:
length ≥ length for F
formula space ≥ variablespace for F
x1 ∨ x2
x1 ∨ x2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
x1 ∨ x2 ∨ y1 ∨ y2
y1 ∨ y2
y1 ∨ y2
Prove that this is (sort of) best one can do for F [⊕]!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 28 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Sketch of Proof of Substitution Theorem
Given refutation of F [⊕], extract “shadow refutation” of F
XOR formula F [⊕] Original formula F
If XOR blackboard implies e.g.¬(x1 ⊕ x2) ∨ (y1 ⊕ y2). . .
write x ∨ y on shadow black-board
For consecutive XOR black-board configurations. . .
can get between correspond-ing shadow blackboards by le-gal derivation steps
. . . (sort of) upper-bounded byXOR derivation length
Length of shadow blackboardderivation . . .
. . . is at most # clauses onXOR blackboard
# variables mentioned onshadow blackboard. . .
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 29 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Pieces Together: Substitution + Pebbling Formulas
Making variable substitutions in pebbling formulas
lifts lower bound from variable space to formula space
maintains upper bound in terms of total space and length
Substitution with XOR over k + 1 variables works against k-DNFresolution
Get our results by
using known pebbling results from literature of 70s and 80s
proving a couple of new pebbling results [Nordstrom ’10]
to get tight trade-offs, showing that resolution can sometimesdo better than black-only pebbling [Nordstrom ’10]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 30 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Pieces Together: Substitution + Pebbling Formulas
Making variable substitutions in pebbling formulas
lifts lower bound from variable space to formula space
maintains upper bound in terms of total space and length
Substitution with XOR over k + 1 variables works against k-DNFresolution
Get our results by
using known pebbling results from literature of 70s and 80s
proving a couple of new pebbling results [Nordstrom ’10]
to get tight trade-offs, showing that resolution can sometimesdo better than black-only pebbling [Nordstrom ’10]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 30 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Pebble Games and Pebbling ContradictionsSubstitution TheoremPutting the Pieces Together
Pieces Together: Substitution + Pebbling Formulas
Making variable substitutions in pebbling formulas
lifts lower bound from variable space to formula space
maintains upper bound in terms of total space and length
Substitution with XOR over k + 1 variables works against k-DNFresolution
Get our results by
using known pebbling results from literature of 70s and 80s
proving a couple of new pebbling results [Nordstrom ’10]
to get tight trade-offs, showing that resolution can sometimesdo better than black-only pebbling [Nordstrom ’10]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 30 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Lower Bounds on Total Space?
Open Question
Are there polynomial-size k-CNF formulas with total refutationspace Ω
((size of F )2
)? in resolution
Answer conjectured to be “yes” by [Alekhnovich et al. 2000]
Or can one at least prove a superlinear lower bound measured in# variables?
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 31 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Stronger Length-Space Trade-offs than from Pebbling?
Open Question
Are there superpolynomial trade-offs in resolution for formulasrefutable in constant space?
Open Question
Are there formulas with trade-offs in the range space > formulasize? Or can every proof be carried out in at most linear space?
Pebbling formulas cannot answer these questions — can impossiblyhave such strong trade-offs
Expecting to hear some exciting news on 2nd question later today!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 32 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Stronger Length-Space Trade-offs than from Pebbling?
Open Question
Are there superpolynomial trade-offs in resolution for formulasrefutable in constant space?
Open Question
Are there formulas with trade-offs in the range space > formulasize? Or can every proof be carried out in at most linear space?
Pebbling formulas cannot answer these questions — can impossiblyhave such strong trade-offs
Expecting to hear some exciting news on 2nd question later today!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 32 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Trade-offs for Stronger Proof Systems?
Recall key technical theorem: amplify space lower bounds throughvariable substitution
Almost completely oblivious to which proof system is being studied
Extended to strictly stronger k-DNF resolution proof systems —maybe can be made to work for other stronger systems as well?
Open Question
Can the Substitution Theorem be proven for, say, Cutting Planesor Polynomial Calculus (with/without Resolution), thus yieldingtime-space trade-offs for these proof systems as well?
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 33 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Trade-offs for Stronger Proof Systems?
Recall key technical theorem: amplify space lower bounds throughvariable substitution
Almost completely oblivious to which proof system is being studied
Extended to strictly stronger k-DNF resolution proof systems —maybe can be made to work for other stronger systems as well?
Open Question
Can the Substitution Theorem be proven for, say, Cutting Planesor Polynomial Calculus (with/without Resolution), thus yieldingtime-space trade-offs for these proof systems as well?
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 33 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Some Related Very Recent Developments
Theorem (Huynh & Nordstrom, Sep ’11)
There are k-CNF formulas refutable in resolution in length O(n)such that any
PCR refutation in length L and monomial space s has
s log L = Ω(
4√
n/ log n)
Cutting Planes refutation in length L and line space s has
s log L = Ω
(6
√n/ log13 n
)
Doesn’t use substitution theorem, but lifting + communicationcomplexity a la [Beame, Huynh & Pitassi ’10]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 34 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Some Related Very Recent Developments
Theorem (Huynh & Nordstrom, Sep ’11)
There are k-CNF formulas refutable in resolution in length O(n)such that any
PCR refutation in length L and monomial space s has
s log L = Ω(
4√
n/ log n)
Cutting Planes refutation in length L and line space s has
s log L = Ω
(6
√n/ log13 n
)
Doesn’t use substitution theorem, but lifting + communicationcomplexity a la [Beame, Huynh & Pitassi ’10]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 34 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Some Related Very Recent Developments
Theorem (Huynh & Nordstrom, Sep ’11)
There are k-CNF formulas refutable in resolution in length O(n)such that any
PCR refutation in length L and monomial space s has
s log L = Ω(
4√
n/ log n)
Cutting Planes refutation in length L and line space s has
s log L = Ω
(6
√n/ log13 n
)
Doesn’t use substitution theorem, but lifting + communicationcomplexity a la [Beame, Huynh & Pitassi ’10]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 34 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Some Related Very Recent Developments
Theorem (Huynh & Nordstrom, Sep ’11)
There are k-CNF formulas refutable in resolution in length O(n)such that any
PCR refutation in length L and monomial space s has
s log L = Ω(
4√
n/ log n)
Cutting Planes refutation in length L and line space s has
s log L = Ω
(6
√n/ log13 n
)
Doesn’t use substitution theorem, but lifting + communicationcomplexity a la [Beame, Huynh & Pitassi ’10]
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 34 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Is Tractability Captured by Space Complexity?
Open Question
Do our theoretical trade-offs show up in real life for state-of-the-artSAT solvers run on pebbling contradictions?
That is, does space complexity capture hardness?
Space suggested as hardness measure in [Ansotegui et al.’08]
Preliminary experiments seem to indicate that pebbling formulasmight actually be hard for SAT solvers
Note that pebbling formulas are always extremely easy with respectto length, so hardness in practice would be intriguing
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 35 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Is Tractability Captured by Space Complexity?
Open Question
Do our theoretical trade-offs show up in real life for state-of-the-artSAT solvers run on pebbling contradictions?
That is, does space complexity capture hardness?
Space suggested as hardness measure in [Ansotegui et al.’08]
Preliminary experiments seem to indicate that pebbling formulasmight actually be hard for SAT solvers
Note that pebbling formulas are always extremely easy with respectto length, so hardness in practice would be intriguing
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 35 / 37
Resolution-Based Proof SystemsOutline of Proofs
Open Problems
Space in ResolutionSpace in Stronger Proof SystemsSpace and SAT solving
Is Tractability Captured by Space Complexity?
Open Question
Do our theoretical trade-offs show up in real life for state-of-the-artSAT solvers run on pebbling contradictions?
That is, does space complexity capture hardness?
Space suggested as hardness measure in [Ansotegui et al.’08]
Preliminary experiments seem to indicate that pebbling formulasmight actually be hard for SAT solvers
Note that pebbling formulas are always extremely easy with respectto length, so hardness in practice would be intriguing
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 35 / 37
Take-Home Message
Modern SAT solvers, although based on old and simple DPLLmethod, can be enormously successful in practice
Key issue is to minimize time and memory consumption
However, our results suggest strong time-space trade-offs thatshould make this impossible
Many remaining open questions about space in proofcomplexity
Main open practical question: is tractability captured by spacecomplexity?
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 36 / 37
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I’m hiring!
2 PhD students
1 postdoc
Starting date autumn 2012
More info soon at www.csc.kth.se/∼jakobn
Thank you for your attention!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 37 / 37
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I’m hiring!
2 PhD students
1 postdoc
Starting date autumn 2012
More info soon at www.csc.kth.se/∼jakobn
Thank you for your attention!
Jakob Nordstrom (KTH) Understanding the Hardness of Proving Formulas Banff 2011 37 / 37