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 1 York University- MATH 1090 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 York University- MATH 1090 06-Deduction 2

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) York University- MATH 1090 06-Deduction 3

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) York University- MATH 1090 06-Deduction 4

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 York University- MATH 1090 06-Deduction 5

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 York University- MATH 1090 06-Deduction 6

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  York University- MATH 1090 06-Deduction 7

Resolution (3) Split  6) A  B ⊢ A and A  B ⊢ B  Split 7) Use Cut Rule to prove  A  B,  A  C ⊢ B  C  Cut Rule A,  A ⊢   York University- MATH 1090 06-Deduction 8