On Propositional QBF Expansions andQ-Resolution
Mikolas Janota1 Joao Marques-Silva1,2
1 INESC-ID/IST, Lisbon, Portugal2 CASL/CSI, University College Dublin, Ireland
SAT 2013, July 8-12
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 1 / 15
QBF Solving
Other
DPLL-Based
QuBE
depQBF
GhostQ
CirQitExpansion-
Based
quantor nenofex
sKizzo
CarefulExpansion
AReQS
RAReQS
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 2 / 15
Quantified Boolean Formula (QBF)
• an extension of SAT with quantifiers
Example
∀y1y2∃x1x2. (y1 ∨ x1) ∧ (y2 ∨ x2)
• we consider prenex form with maximal blocks of variables
∀U1∃E2 . . . ∀U2N−1∃E2N . φ
Solving
• DPLL — Q-Resolution (QuBE, depqbf, etc.)
• Expansion — ?? (Quantor, sKizzo, Nenofex)• “Careful” expansion (AReQS,RAReQS)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 3 / 15
Quantified Boolean Formula (QBF)
• an extension of SAT with quantifiers
Example
∀y1y2∃x1x2. (y1 ∨ x1) ∧ (y2 ∨ x2)
• we consider prenex form with maximal blocks of variables
∀U1∃E2 . . . ∀U2N−1∃E2N . φ
Solving
• DPLL — Q-Resolution (QuBE, depqbf, etc.)
• Expansion — ?? (Quantor, sKizzo, Nenofex)• “Careful” expansion (AReQS,RAReQS)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 3 / 15
Quantified Boolean Formula (QBF)
• an extension of SAT with quantifiers
Example
∀y1y2∃x1x2. (y1 ∨ x1) ∧ (y2 ∨ x2)
• we consider prenex form with maximal blocks of variables
∀U1∃E2 . . . ∀U2N−1∃E2N . φ
Solving
• DPLL — Q-Resolution (QuBE, depqbf, etc.)
• Expansion — ?? (Quantor, sKizzo, Nenofex)• “Careful” expansion (AReQS,RAReQS)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 3 / 15
Quantified Boolean Formula (QBF)
• an extension of SAT with quantifiers
Example
∀y1y2∃x1x2. (y1 ∨ x1) ∧ (y2 ∨ x2)
• we consider prenex form with maximal blocks of variables
∀U1∃E2 . . . ∀U2N−1∃E2N . φ
Solving
• DPLL — Q-Resolution (QuBE, depqbf, etc.)
• Expansion — ?? (Quantor, sKizzo, Nenofex)• “Careful” expansion (AReQS,RAReQS)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 3 / 15
Q-resolution
Q-resolution = Q-resolution rule+∀-reduction
Q-resolution ruleC1, C2 with one and only one complementary literal l , where l isexistential
• derive C1 ∪C2 r {l , l}
∀-reduction
• if k ∈ C is universal with highest level in C , remove k from C
Tautologous resolvents are generally unsound!
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 4 / 15
Q-resolution
Q-resolution = Q-resolution rule+∀-reduction
Q-resolution ruleC1, C2 with one and only one complementary literal l , where l isexistential
• derive C1 ∪C2 r {l , l}
∀-reduction
• if k ∈ C is universal with highest level in C , remove k from C
Tautologous resolvents are generally unsound!
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 4 / 15
Q-resolution
Q-resolution = Q-resolution rule+∀-reduction
Q-resolution ruleC1, C2 with one and only one complementary literal l , where l isexistential
• derive C1 ∪C2 r {l , l}
∀-reduction
• if k ∈ C is universal with highest level in C , remove k from C
Tautologous resolvents are generally unsound!
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 4 / 15
Q-resolution
Q-resolution = Q-resolution rule+∀-reduction
Q-resolution ruleC1, C2 with one and only one complementary literal l , where l isexistential
• derive C1 ∪C2 r {l , l}
∀-reduction
• if k ∈ C is universal with highest level in C , remove k from C
Tautologous resolvents are generally unsound!
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 4 / 15
Expansion
∀x . Φ = Φ[x/0] ∧ Φ[x/1]
∃x . Φ = Φ[x/0] ∨ Φ[x/1]
Fresh variables in order to keep prenex form
∃e1∀u2∃e3. (e1 ∨ e3) ∧ (e3 ∨ e1) ∧ (u2 ∨ e3) ∧ (u2 ∨ e3)
∃e1eu2/03 eu2/13 . (e1 ∨ e
u2/03 ) ∧ (e
u2/03 ∨ e1) ∧
(e1 ∨ eu2/13 ) ∧ (e
u2/13 ∨ e1) ∧
eu2/03 ∧eu2/13
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 5 / 15
Expansion
∀x . Φ = Φ[x/0] ∧ Φ[x/1]
∃x . Φ = Φ[x/0] ∨ Φ[x/1]
Fresh variables in order to keep prenex form
∃e1∀u2∃e3. (e1 ∨ e3) ∧ (e3 ∨ e1) ∧ (u2 ∨ e3) ∧ (u2 ∨ e3)
∃e1eu2/03 eu2/13 . (e1 ∨ e
u2/03 ) ∧ (e
u2/03 ∨ e1) ∧
(e1 ∨ eu2/13 ) ∧ (e
u2/13 ∨ e1) ∧
eu2/03 ∧eu2/13
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 5 / 15
Expansion
∀x . Φ = Φ[x/0] ∧ Φ[x/1]
∃x . Φ = Φ[x/0] ∨ Φ[x/1]
Fresh variables in order to keep prenex form
∃e1∀u2∃e3. (e1 ∨ e3) ∧ (e3 ∨ e1) ∧ (u2 ∨ e3) ∧ (u2 ∨ e3)
∃e1eu2/03 eu2/13 . (e1 ∨ e
u2/03 ) ∧ (e
u2/03 ∨ e1) ∧
(e1 ∨ eu2/13 ) ∧ (e
u2/13 ∨ e1) ∧
eu2/03 ∧eu2/13
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 5 / 15
How to prove by expansion?
1. Expand all universal variables
2. Refute by propositional resolution
Why only universals?
1. conjunction of CNF is still CNF
2. ∃-expansion “doing the work of resolution”
Partial Expansions
Only certain values may be needed:
∀u∃e. (u ∨ e) ∧ (u ∨ e) ∧ (u ∨ e) (false)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 6 / 15
How to prove by expansion?
1. Expand all universal variables
2. Refute by propositional resolution
Why only universals?
1. conjunction of CNF is still CNF
2. ∃-expansion “doing the work of resolution”
Partial Expansions
Only certain values may be needed:
∀u∃e. (u ∨ e) ∧ (u ∨ e) ∧ (u ∨ e) (false)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 6 / 15
How to prove by expansion?
1. Expand all universal variables
2. Refute by propositional resolution
Why only universals?
1. conjunction of CNF is still CNF
2. ∃-expansion “doing the work of resolution”
Partial Expansions
Only certain values may be needed:
∀u∃e. (u ∨ e) ∧ (u ∨ e) ∧ (u ∨ e) (false)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 6 / 15
How to prove by expansion?
1. Expand all universal variables
2. Refute by propositional resolution
Why only universals?
1. conjunction of CNF is still CNF
2. ∃-expansion “doing the work of resolution”
Partial Expansions
Only certain values may be needed:
∀u∃e. (u ∨ e) ∧ (u ∨ e) ∧ (u ∨ e) (false)
∃eu/0eu/1. eu/0 ∧ eu/0 ∧ eu/1 (false full)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 6 / 15
How to prove by expansion?
1. Expand all universal variables
2. Refute by propositional resolution
Why only universals?
1. conjunction of CNF is still CNF
2. ∃-expansion “doing the work of resolution”
Partial Expansions
Only certain values may be needed:
∀u∃e. (u ∨ e) ∧ (u ∨ e) ∧ (u ∨ e) (false)
∃eu/0eu/1. eu/0 ∧ eu/0 ∧ eu/1 (false full)
∃eu/0. eu/0 ∧ eu/0 (false partial)
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 6 / 15
Recursive Partial Expansion
∀U1
∃E2
∀U3
∃E4
φ
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 7 / 15
Recursive Partial Expansion
∧. . .∃E2 ∃E2
∀U3
∃E4
φ
∀U3
∃E4
φ
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 7 / 15
Recursive Partial Expansion
∧. . .∃E2 ∃E2∧
. . . ∃E4
φ
∃E4
φ
∧. . .∃E4
φ
∃E4
φ
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 7 / 15
∀Exp+Res
Proof: (T , π)
(1) Expansion tree T : for each block of variables it tells us how toexpand it.
T
u2/1
u1/0
u2/0 u2/1
u1/1
(2) Propositional Resolution Refutation π of expansion resultingfrom the expansion tree T .
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 8 / 15
∀Exp+Res
Proof: (T , π)(1) Expansion tree T : for each block of variables it tells us how toexpand it.
T
u2/1
u1/0
u2/0 u2/1
u1/1
(2) Propositional Resolution Refutation π of expansion resultingfrom the expansion tree T .
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 8 / 15
∀Exp+Res
Proof: (T , π)(1) Expansion tree T : for each block of variables it tells us how toexpand it.
T
u2/1
u1/0
u2/0 u2/1
u1/1
(2) Propositional Resolution Refutation π of expansion resultingfrom the expansion tree T .
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 8 / 15
Performing Expansion
• For a clause C = ei ∨ u ∨ ek , for τ = τ1, . . . , τn
E (τ1, . . . , τn, C ) = eτ1,...,τi/2i ∨ e
τ1,...,τk/2k if u[τ ] = 0
E (τ1, . . . , τn, C ) = 1 if u[τ ] = 1
• For an expansion tree T and a matrix φ consider the union ofclauses E (τ,C ) for all branches τ ∈ T and C ∈ φ.
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 9 / 15
Performing Expansion
• For a clause C = ei ∨ u ∨ ek , for τ = τ1, . . . , τn
E (τ1, . . . , τn, C ) = eτ1,...,τi/2i ∨ e
τ1,...,τk/2k if u[τ ] = 0
E (τ1, . . . , τn, C ) = 1 if u[τ ] = 1
• For an expansion tree T and a matrix φ consider the union ofclauses E (τ,C ) for all branches τ ∈ T and C ∈ φ.
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 9 / 15
From Tree Q-resolution to ∀Exp+Res
C1 ∨ C2
x ∨ C1 x ∨ C2
x ∈ D1 x , y ∈ D2
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 10 / 15
From Tree Q-resolution to ∀Exp+Res
C ′1 ∨ C ′2
xτ1,...,τj ∨ C ′1 xτ1,...,τj ∨ C ′2
xτ1,...,τj ∈ E (D1, τ) xµ1,...,µj , yµ1,...,µk ∈ E (D1, µ)
µ1, . . . , µj = τ1, . . . , τj
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 10 / 15
From Tree Q-resolution to ∀Exp+Res
C ′1 ∨ C ′2
xτ1,...,τj ∨ C ′1 xτ1,...,τj ∨ C ′2
xτ1,...,τj ∈ E (D1, τ) xµ1,...,µj , yµ1,...,µk ∈ E (D1, µ)
µ1, . . . , µj = τ1, . . . , τj
C3 ∨ C4
y ∨ C3 y ∨ C4
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 10 / 15
From Tree Q-resolution to ∀Exp+Res
C ′1 ∨ C ′2
xτ1,...,τj ∨ C ′1 xτ1,...,τj ∨ C ′2
xτ1,...,τj ∈ E (D1, τ) xµ1,...,µj , yµ1,...,µk ∈ E (D1, µ)
µ1, . . . , µj = τ1, . . . , τj
C3 ∨ C4
y ∨ C3 y ∨ C4
y ∈ D3
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 10 / 15
From Tree Q-resolution to ∀Exp+Res
C ′1 ∨ C ′2
xτ1,...,τj ∨ C ′1 xτ1,...,τj ∨ C ′2
xτ1,...,τj ∈ E (D1, τ) xµ1,...,µj , yµ1,...,µk ∈ E (D1, µ)
µ1, . . . , µj = τ1, . . . , τj
C3 ∨ C4
yρ1,...,ρk ∨ C ′3 yµ1,...,µk ∨ C ′4
yρ1,...,ρk ∈ E (D3, ρ)
ρ1, . . . , ρk = µ1, . . . , µk
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 10 / 15
From Expansion Refutation to Q-resolution
• Why don’t we just revert substitutions?
xτ xτ
⊥
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 11 / 15
From Expansion Refutation to Q-resolution
• Why don’t we just revert substitutions?
xτ xτ
⊥
u ∨ x u ∨ x
u
⊥
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 11 / 15
From Expansion Refutation to Q-resolution
• Why don’t we just revert substitutions?
xτ xτ
⊥
u ∨ x u ∨ x
u
⊥
xτ ∨ yµ xτ ∨ yρ
yρ ∨ yµ
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 11 / 15
From Expansion Refutation to Q-resolution
• Why don’t we just revert substitutions?
xτ xτ
⊥
u ∨ x u ∨ x
u
⊥
xτ ∨ yµ xτ ∨ yρ
yρ ∨ yµ
x ∨ u ∨ y x ∨ u ∨ y
u ∨ u ∨ y ∨ y
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 11 / 15
From Expansion Refutation to Q-resolution
• Why don’t we just revert substitutions?
xτ xτ
⊥
u ∨ x u ∨ x
u
⊥
xτ ∨ yµ xτ ∨ yρ
yρ ∨ yµ
x ∨ u ∨ y x ∨ u ∨ y
u ∨ u ∨ y ∨ y
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 11 / 15
From Expansion Refutation to Q-resolution
• Why don’t we just revert substitutions?
xτ xτ
⊥
u ∨ x u ∨ x
u
⊥
xτ ∨ yµ xτ ∨ yρ
yρ ∨ yµ
x ∨ u ∨ y x ∨ u ∨ y
u ∨ u ∨ y ∨ y
• Such a construction is possible if propositional resolutionfollows the order of the prefix, starting with the innermostlevels.
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 11 / 15
What is hard for ∀Exp+Res
⊥
u1 ∨ e2
u1 ∨ e2 ∨ u3 ∨ u4
u1 ∨ e2 ∨ u3 ∨ u4 ∨ e5u3 ∨ e5
u1 ∨ e2
u1 ∨ e2 ∨ u3 ∨ u4
u1 ∨ e2 ∨ u3 ∨ u4 ∨ e5
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 12 / 15
What Seems to Be Hard for Q-resolution
xi ∨ z ∨ C1i xi ∨ z ∨ C2
i
x1 ∨ z ∨ y1 x1 ∨ z ∨ y1x2 ∨ z ∨ y1 x2 ∨ z ∨ y1x3 ∨ z ∨ y1 x3 ∨ z ∨ y1x4 ∨ z ∨ y1 x4 ∨ z ∨ y1
z/0 z/1
x1 ∨ yz/01 x1 ∨ y
z/11
x2 ∨ yz/01 x2 ∨ y
z/11
x3 ∨ yz/01 x3 ∨ y
z/11
x4 ∨ yz/01 x4 ∨ y
z/11
Figure : Example formula for n = 1
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 13 / 15
Summary and Future Work
• We have defined a proof system based on “careful”expansions and propositional resolution.
• Such system simulates tree Q-resolution.
• Q-resolution can simulate a fragment of this system, whenvariables are resolved “inside out”.
• We conjecture that the systems are incomparable. Showingsuch is the subject of future work.
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 14 / 15
Summary and Future Work
• We have defined a proof system based on “careful”expansions and propositional resolution.
• Such system simulates tree Q-resolution.
• Q-resolution can simulate a fragment of this system, whenvariables are resolved “inside out”.
• We conjecture that the systems are incomparable. Showingsuch is the subject of future work.
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 14 / 15
Summary and Future Work
• We have defined a proof system based on “careful”expansions and propositional resolution.
• Such system simulates tree Q-resolution.
• Q-resolution can simulate a fragment of this system, whenvariables are resolved “inside out”.
• We conjecture that the systems are incomparable. Showingsuch is the subject of future work.
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 14 / 15
Summary and Future Work
• We have defined a proof system based on “careful”expansions and propositional resolution.
• Such system simulates tree Q-resolution.
• Q-resolution can simulate a fragment of this system, whenvariables are resolved “inside out”.
• We conjecture that the systems are incomparable. Showingsuch is the subject of future work.
Janota and Marques-Silva On Propositional QBF Expansions and Q-Resolution 14 / 15