6- Deduction theorem & Resolution Ref: G. Tourlakis, Mathematical - - PowerPoint PPT Presentation

SC/MATH 1090

6- Deduction theorem & Resolution

Ref: G. Tourlakis, Mathematical Logic, John Wiley & Sons, 2008.

York University

Department of Computer Science and Engineering

York University- MATH 1090

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06-Deduction

Overview

• Deduction Theorem

– Proving a Lemma (by induction) to prove deduction theorem (by induction on the length of proof!) – Why is deduction theorem important?

• Resolution- Easy life!!!

– Definitions – Procedure for resolution – Examples

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The Deduction Theorem

• Lemma. A (B  C) ⊢ A  (D[p:=B]  D[p:=C])
• Metatheorem. (Deduction Theorem) If 

{A} ⊢B, then also  ⊢A  B.

• Corollary. +A ⊢B iff  ⊢A  B.
• Why is it so important?

– Theory: All theorems are equivalent to absolute theorems. – Logic users: when proving  ⊢A  B, instead prove +A ⊢B (which is usually easier)

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

Equivalent statements

• Metatheorem. The following are equivalent:

1)  ⊢  2) For all A,  ⊢A 3) For some B, we have  ⊢ B   B  is called inconsistent if any of above is proven for .

• Corollary. (Proof by Contradiction)  ⊢ A iff  +  A ⊢ .

– Logic users: when proving  ⊢A , instead prove +  A ⊢  (prove inconsistency if  A is assumed)

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Definitions

• Definition. (Literal) Atomic formulae and their negation,

also formula variables, and their negation are literals.

• Examples of literals:

p, p,   , A,  B

• Definition. (Clause) A clause is a disjunction of literals.
• Examples of clauses:

p  p ,    A   B

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Resolution (1)

1) Using Deduction and proof by contradiction to move everything to the assumption side, and have  on the conclusion side

• If +A ⊢B then  ⊢A  B.

Deduction Theorem

• If  +  A ⊢  then  ⊢ A.

Proof by contradiction

In other words, instead of proving  ⊢A  B, it is sufficient to prove  + {A ,  B} ⊢ . 2) Remove any  using Ping-pong

• A  B  (A  B)  (B  A)

Ping-pong theorem

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Resolution (2)

3) Remove any  using implication theorem

• A  B  A  B

Implication theorem

4) Move negation inwards using de Morgan

• (A  B)   A   B

de Morgan 1

• (A  B)   A   B

de Morgan 2

5) Distribute  over 

• A  (B  C)  (A  B)  (A  C) Distributivity of  over 

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Resolution (3)

6) Split 

• A B ⊢ A and A B ⊢ B

Split

7) Use Cut Rule to prove 

• AB, AC ⊢ BC

Cut Rule

• A, A ⊢

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