Deductive Systems

Deductive Systems

Lecture - 3

Axiomatic System

Prof Saroj Kaushik, CSE, IITD

Axiomatic System (AS) for PL

● AS is based on the set of only three axioms and

one rule of deduction.

− It is minimal in structure but as powerful as the truth table

and natural deduction approaches.

− The proofs of the theorems are often difficult and require a

guess in selection of appropriate axiom(s) and rules.

− These methods basically require forward chaining strategy

where we start with the given hypotheses and prove the

goal.


Axiomatic System for PL -Cont

● Three axioms and one rule of deduction.

Axiom1 (A1): α → (β → α)

Axiom2 (A2): (α →(β→γ)) →((α → β) → (α → γ))

Axiom3 (A3): (~ α → ~ β) → ( β → α)

Modus Ponen (MP) defined as follows:

Hypotheses: α → β and α Consequent: β


Cont…

Definition: A deduction of a formula in Axiomatic

System for Propositional Logic is a sequence of well-

formed formulae α1, α2, ..., αn such that for each i,

(1≤ i ≤ n), either

– Either αi is an axiom or αi is a hypothesis (given to be true)

– Or αi is derived from αj and αk where j, k < i using modus ponen inference rule.

� We call αi to be a deductive consequence of {α1, ...,αi-1 }.

� It is denoted by {αααα1, .. , ααααi-1 } |- ααααi. More formally, deductive consequence is defined on next slide.


Cont…

Definition: If ∑ is a set of hypotheses involved in

the deduction of α as defined above, then α is called to be a deductive consequence of ∑ or α is

deducible from ∑ . It is written as ∑ |-α.

Definition: If ∑ is an empty set and α is deduced,

then we write |- α. In this case α is deduced from axioms only and no hypotheses are used. In such

case we call α to be a theorem.

� Instances of the axioms are the simplest forms of

theorems in Axiomatic Theory.


Cont…

� It should be noted that proofs need not be unique.

� We have to carefully select axioms, which can lead to the proof.

� The proof of a theorem can be tricky and hard.

� The proofs require mixture of bottom up and top

down reasoning.

� Start reasoning bottom up seeing what we can do

with axioms and top down reasoning seeing what

might possibly lead to answer.


Examples

Establish the following:

Ex1:{Q} |- (P→Q) i.e.,P→Q is a deductive consequence of {Q}.

{Hypothesis} Q (1)

{Axiom A1} Q → (P → Q) (2)

{MP, (1,2)} P →→→→ Q proved


Examples – Cont…

Ex2:

{ P → Q, Q → R } |- ( P → R ) i.e., P → R is a

deductive consequence of { P → Q, Q → R }.

{Hypothesis} P → Q (1)

{Hypothesis} Q → R (2)

{Axiom A1} (Q→ R) → (P → (Q → R)) (3)

{MP, (2, 3)} P → (Q → R) (4)

{Axiom A2} (P → (Q → R)) →

((P → Q) → (P → R)) (5)

{MP , (4, 5)} (P → Q) → (P → R) (6)

{MP, (1, 6)} P →→→→ R proved


Deduction Theorems in AS

Deduction Theorem:

If ∑ is a set of hypotheses and α and β are

well-formed formulae , then {∑ ∪ α } |- βimplies ∑ |- (α → β ).

Converse of deduction theorem:

Given ∑ |- (α → β ),

we can prove { ∑ ∪ α } |- β.


Useful Tips

1. Given α, we can easily prove β → α for any well-formed formulae α and β.

2. If α → β is to be proved, then include α in the set of hypotheses ∑ and derive β from the set {∑ ∪ α}. Then using deduction theorem, we conclude α → β.


Examples

Example1:

Prove |- ~P →→→→ (P →→→→ Q) using deduction theorem of Axiomatic System.

Proof: Prove {~ P} |- (P → Q) and |- ~ P→(P→Q) follows from deduction theorem.


Proof

{Prove} {~ P} |- (P →→→→ Q)

{Hypothesis} ~ P (1)

{Axiom A1} ~ P → (~ Q → ~ P) (2)

{MP, (1, 2)} ~ Q → ~ P (3)

{Axiom A3} (~ Q → ~ P) → ( P → Q) (4)

{MP, (3,4)} P →→→→ Q proved

Hence by using deduction theorem, we conclude

|- ~ P → (P → Q)


Cont…

Example2:Prove that |- (~ ~ P → ~ ~P) is a theorem.

Proof:{Theorem} |- (~ ~P →→→→ ~ ~P){Axiom A1} ~ ~ P → (( Q → ~ ~P)

→ ~ ~P) (1){Axiom A2} (~ ~P → (( Q → ~ ~P) → ~ ~P))

→ ((~ ~P → ( Q → ~ ~ P)) → (~ ~P → ~ ~P)) (2)

{ MP, (1, 2)} ((~ ~P → ( Q → ~ ~ P)) → (~ ~P → ~ ~P)) (3)

{Axiom A1} ~ ~P → ( Q → ~ ~ P) (4){ MP, (3, 4) } ~ ~P →→→→ ~ ~ P proved


Cont…

Definition: A truth valuation is a function ν from a set of well-formed formulae to the set {T, F} such

that for any well-formed formulae α and β

– ν(~ α ) ≠ ν (α )

– ν (α → β ) = F iff ν (α ) = T and ν (β ) = F

� The truth assignment to the atoms uniquely

determines the truth valuation of all formulae.

Definition: A formula α is said to be valid (denoted

by |= α) if and only if for all valuation ν, ν(α) = T.

Lemma 1: If α and β are well-formed formulae,

then |= α and |= (α → β) implies |= β.


Cont…

Definition: A formula α is said to be inconsistent if

and only if for all valuation ν, ν(α) = F.

Definition: If α is a formula and ∃ a valuation ν such

that ν(α) = T, then α is said to be consistent.

Definition: A set of formulae is said to be mutually

consistent if and only if they are all true

simultaneously for some valuation.

Definition: A set of formulae is said to be mutually

inconsistent if and only if there exists no valuation

under which conjunction of formulae is true.


Soundness and Completeness in AS

Theorem : If α is a formula in AS, then α is a theorem if

and only if α is valid.

� (Soundness): Every theorem is valid i.e.,

|- αααα →→→→ |= αααα.

� (Completeness): if α is valid then α is a theorem

i.e., |= αααα →→→→ |- αααα.

� (Consistency): Axiomatic System for propositional

logic is consistent if it is not possible to prove both αand ~ α for any well-formed formula α i.e., not

both |- α and |- ~ α.

Semantic Tableaux Method


Semantic Tableaux System in PL

● Earlier approaches require

− construction of proof of a formula from given set of

formulae and are called direct methods.

● In semantic tableaux,

− the set of rules are applied systematically on a formula or

set of formulae to establish its consistency or

inconsistency.

● Semantic tableau

− binary tree constructed by using semantic rules with a

formula as a root

● Assume α and β be any two formulae.


Semantic Tableaux Rules

• Let α and β be any two formulae. Rule 1: A tableau for a formula (α Λ β) is constructed by

adding both α and β to the same path (branch). This can be represented as follows:

α Λ βαβ

Interpretation: α Λ β is true if both α and β are true


Rules - Cont…

Rule 2: A tableau for a formula ~ (α Λ β) is constructed by adding two alternative paths one containing ~ α and other containing ~ β

~ (α Λ β)

~ α ~ β

Interpretation: ~ (α Λ β ) is true if either ~ αor ~ β is true


Cont…

Rule 3: A tableau for a formula (α V β) is constructed by adding two new paths one containing α and other containing β.

α V β

α β

Interpretation: α V β is true if either αor β is true


Cont…

Rule 4: A tableau for a formula ~ (α V β) is

constructed by adding both ~ α and ~ β to the same path. This can be expressed as follows:

~ ( α V β)

~ α

~ β

Rule 5: Semantic tableau for ~~ α~~ α

α


Rule 6: Semantic tableau for α → β

α → β

~ α β

Rule 7: Semantic tableau for ~ ( α → β)

~ (α → β)

α

~ β


Rule 8: Semantic tableau for α ↔ β α↔ β ≅ (α Λ β) V (~ α Λ ~ β)

α ↔ β

α Λ β ~ α Λ ~ β

Rule 9: Semantic tableau for ~ (α ↔ β)

~ (α ↔ β) ≅ (α Λ ~ β) V (~ α Λ β)~ (α ↔ β)

α Λ ~ β ~ α Λ β


Consistency and Inconsistency

� If an atom P and ~ P appear on a same

path of a semantic tableau, – then inconsistency is indicated and such path is

said to be contradictory or closed (finished)

path.

− Even if one path remains non contradictory or

unclosed (open), then the formula α at the root of a tableau is consistent.


Valuation

� A valuation ν is said to be a model of α (or νsatisfies α) iff ν (α) = T.

� In tableaux approach, model for a consistent

formula α is constructed as follows:

– On an open path, assign truth values to atoms

(positive or negative) of α which is at the root of a

tableau such that α is made to be true.

– It is achieved by assigning truth value T to each

atomic formula (positive or negative) on that path.


Contradictory Tableau

• Contradictory tableau (or finished tableau) is defined to be a tableau in which all the paths are contradictory or closed

(finished).

• If a tableau for a formula α at the root is a contradictory tableau, then a formula α is said to be inconsistent.

• A formula α is consistent if there is at least on open path in a tableau with root α


Example – Inconsistent

Example: Show that

α : (P Λ Q → R) Λ (~P → S) Λ Q Λ ~ R Λ ~ S is inconsistent using tableaux method.

{T-root} (P Λ Q → R) Λ ( ~P → S) Λ Q Λ ~ R Λ ~ S (1)

{Apply rule 1 to 1} P Λ Q → R (2)

~P → S (3)

Q

~ R

~ S

{Apply rule 6 to 3} P S

Closed: {S, ~ S} on the path

{Apply rule 6 to 2)} ~ (P Λ Q) R

Closed { R, ~ R}

~P ~ Q

Closed {P, ~ P} Closed{~ Q, Q}


Example - Consistent

Problem: Show that α: ( Q Λ~R) Λ (R → P) is consistent and find its model.

Solution:{T-root} ( Q Λ ~ R) Λ ( R → P) (1)

{Apply rule 1 to 1} (Q Λ ~ R) (2)

( R → P) (3)

{Apply rule 1 to 2} Q

{Apply rule 6 to 3} ~R

~ R P

open open


Example -Cont…

� Since tableau for α has open paths, we conclude

that α is consistent.

� The models are constructed by assigning T to all

atomic formulae appearing on open paths.

– Assign Q = T and ~ R = T i.e., R = F.

• So { Q = T, R = F } is a model of αααα.

– Assign Q = T and ~ R = T and P = T.

• So { P = T , Q = T, R = F } is another model.

� Useful Tip:

– Thumb rule for constructing a tableau is to

apply non branching rules before the branching

rules in any order


Important Definitions

� A set of formulae {α1, α2, ….,αn } is said to be consistent if the formulae in a set are simultaneously true for some model i.e., if a tableau for α1 Λ α2 Λ ….. Λ αn has at least one open (or non contradictory) path.

� A set of formulae {α1, α2, ….,αn } is said to be inconsistent iff all the formulae can not be true simultaneously i.e., tableau for (α1Λ α2Λ ….Λ αn )as a root is a contradictory tableau.


Definitions – Cont…

� A formula α is tableau provable if tableau with root entry as ~ α is contradictory tableau.

– A tableau proof of a formula α is a contradictory tableau with root entry as ~ α.

� A formula α is valid if α is tableau provable.


Soundness and Completeness

� Theorem: (Soundness)

If α is tableau provable ( |- α ) , then α is

valid ( |= α ) i.e., |- α ⇒ |= α.

� Theorem: (Completeness)

If α is valid, then α is tableau provable i.e.,

|= α ⇒ |- α.


Example - Validity

Example: Show that α : P → ( Q → P) is valid

Solution: In order to show that α is a valid, we will try to show that α is tableau provable i.e., ~ α is inconsistent.

{T-root} ~ (P → ( Q → P)) (1)

{Apply rule7 to 1} P

~ ( Q → P) (2)

{Apply rule 7 to 2} Q

~P

Closed {P, ~ P}

Hence P → ( Q → P) is valid.


Logical Consequence

� A tableau proof of a formula α from a set of premises ∑ = {α1, α2 ,…,αn } is a tableau obtained from (α1 Λ α2 Λ ….. Λ αn) with ~ α as a root entry.

� If it is a contradictory tableau, then we say that α is tableau provable from ∑.– It is denoted by ∑ |- α.

� A formula α is said to be a logical consequence (LC) of a set of premises ∑ , if α is tableau provable from ∑.– It is denoted by ∑ |= α.


Soundness and Completeness of LC

Theorem: (Soundness of deduction from

premises): If there is a tableau proof of αfrom a set of premises ∑, then α is a

logical consequence of ∑, i.e., ∑ |- α ⇒

∑ |= α.

Theorem: (Completeness of deduction from

premises): If α is a logical consequence of a set ∑ of premises, then α is tableau provable from ∑, i.e., ∑ |= α ⇒ ∑ |- α.

Resolution Method in PL


Resolution Refutation in PL

� Resolution refutation is another simple method to prove a formula by contradiction.– Here negation of goal to be proved is added to given set

of clauses.

– It is shown then that there is a refutation in new set using resolution principle.

� Resolution: During this process we need to identify – two clauses, one with positive atom (P) and other with

negative atom (~P) for the application of resolution rule.


Cont…

� Resolution is based on modus poneninference rule.

– This method is most favoured for developing

computer based theorem provers.

– Automatic theorem provers using resolution are simple and efficient systems .

� Resolution is performed on special types of formulae called clauses.

– Clause is propositional formula expressed using

{V, ~ } operators.


Conjunctive and Disjunctive

Normal Forms

� In Disjunctive Normal Form (DNF), – a formula is represented in the form

– (L11 Λ ….. Λ L1n ) V ..… V (Lm1 Λ ….. Λ Lmk ), where all Lij are literals. It is a disjunction

of conjunction.

� In Conjunctive Normal Form (CNF),– a formula is represented in the form

– (L11 V ….. V L1n ) Λ …… Λ (Lp1 V ….. V Lpm ) , where all Lij are literals. It is a

conjunction of disjunction.

� A clause is a special formula expressed as disjunction of literals. If a clause contains only one literal, then it is called unit clause.


Conversion of a Formula to its CNF

� Each formula in Propositional Logic can be easily transformed into its equivalent DNF or CNF representation using equivalence laws .

– Eliminate → and ↔ by using the following equivalence laws.

P → Q ≅ ~ P V Q

P ↔ Q ≅ ( P → Q) Λ ( Q → P)

– Eliminate double negation signs by using

~ ~ P ≅ P


Cont…

� Use De Morgan’s laws to push ~ (negation)

immediately before atomic formula.

~ ( P Λ Q) ≅ ~ P V ~ Q

~ ( P V Q) ≅ ~ P Λ ~ Q

� Use distributive law to get CNF.

P V (Q Λ R) ≅ (P V Q) Λ (P V R)

� We notice that CNF representation of a formula is

of the form

– (C1 Λ….. ΛCn ) , where each Ck , (1≤ k ≤ n ) is a clause that is disjunction of literals.


Resolution of Clauses

� If two clauses C1 and C2 contain a

complementary pair of literals {L, ~L}, then

– these clauses can be resolved together by deleting L from C1 and ~ L from C2 and constructing a new clause by the disjunction of the remaining literals in C1 and C2.

� The new clause thus generated is called

resolvent of C1 and C2 .

– Here C1 and C2 are called parents of resolved clause.

– If the resolvent contains one or more set of complementary pair of literals, then resolvent is always true.


Resolution Tree

� Inverted binary tree is generated with the last node of the binary tree to be a resolvent.

� This also called resolution tree.

Example: Find resolvent of:

C1 = P V Q V R

C2 = ~ Q V ~ W

C3 = ~ P V ~ W


Example- Resolution Tree

P V Q V R ~ Q V ~W

{Q, ~ Q}

P V R V ~W ~ P V ~ W

{P, ~P}

R V ~W

� Thus Resolvent(C1,C2, C3) = R V ~W


Definitions

Theorem: If C is a resolvent of two clauses C1 and

C2, then C is a logical consequence of {C1 , C2 }.

Definition: A deduction of an empty clause from a

set S of clauses is called a resolution refutation of

S.

Theorem: (Soundness & Completeness of

resolution): There is a resolution refutation of S iffS is unsatisfiable / inconsistent.


Cont…

Theorem: Let S be a set of clauses.

� A clause C is a logical consequence of S iff

the set S’= S ∪ {~ C} is unsatisfiable.

� In other words, C is a logical consequence

of a given set S iff an empty clause is

deduced from the set S'.


Example

Example: “Mary will get her degree if she registers

as a student and pass her exam. She has registered herself as a student. She has passed

her exam”. Show that she will get a degree.

Solution: Symbolize above statements as follows:

R: Mary is a registered student

P: Mary has passed her exam

D: Mary gets her degree

� The formulae corresponding to above listed

sentences are as follows:


Cont…

� Mary will get her degree if she registers as a student and pass her exam.

R Λ P → D ≅ (~ R V ~ P V D)

� She has registered herself as a student. R

� She has passed her exam.P

� Conclude “Mary will get a degree”.D


Example – Cont…

� Set of clauses are:

– S = {~ R V ~ P V D , R, P }

� Add negation of "Mary gets her degree (= D)" to S.

� New set S' is:

– S' = {~ R V ~ P V D , R, P, ~ D}

� We can easily see that empty clause is

deduced from above set.

� Hence we can conclude that “Mary gets her degree”


Deriving Contradiction

~ R V ~ P V D R

~ P V D P

D ~ D


Exercises

I. Establish the following:1. { P → Q, Q → R } |- ( P → R )

2. { P → Q} |- (R → P) → (R → Q)

3. { P } |- (~ P → Q) 4. { ~Q, P → ( ~Q → R) } |- P → R

5. {P → Q , ~ Q } |- ~ P. This is called Modus Tollen rule.

II. Prove the following theorems1. |- (P → P)

2. |- (~ P → P) → P

3. |- (P → Q) → (~ Q → ~ P)

4. |- (P → ~ Q) → ( Q → ~ P)

III. Give tableau proof of each of the following formulae and show that formulae are valid.1. P → ( Q → P)

2. (P Λ (Q V R) ↔ (( P Λ Q) V ( P Λ R))

3. ~ (P V Q) ↔ (~ P Λ ~ Q)

IV. Are the following arguments valid?1. If John lives in England then he lives in UK. John lives in England. Therefore, John lives in UK.

2. If John lives in England then he lives in UK. John lives in UK. Therefore, John lives in England.

3. If John lives in England then he lives in UK. John does not live in UK. Therefore, John does not live in England.

V. Prove by resolution refutation 1. {P Λ Q , ~ P V R} |= Q V R

2. { P , Q → R , P → R} |= P Λ R

3. {P → Q Λ R , P} |= R

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