asg3-sol - MATH1090 Problem Set 3 —Solutions November...

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Unformatted text preview: MATH1090 Problem Set 3 —Solutions November 2007 Dept. of MATH and STATS MATH1090. Problem Set 3 —Solutions Posted: Nov. 21, 2007 1. (5 MARKS) Use resolution (in combination with the deduction theorem) —but NOT Post’s theorem— to prove A → B ‘ C ∨ A → C ∨ B Proof. Prove instead A → B,C ∨ A, ¬ ( C ∨ B ) ‘ ⊥ . 2. (5 MARKS) Use resolution (in combination with the deduction theorem) —but NOT Post’s theorem— to prove A → ( B → C ) , B ‘ A → C Page 1 G. Tourlakis MATH1090 Problem Set 3 —Solutions November 2007 Proof. Prove instead A → ( B → C ) , B, A, ¬ C ‘ ⊥ . 3. (5 MARKS) Use resolution (in combination with the deduction theorem) —but NOT Post’s theorem— to prove ‘ ( p → ( q → r )) → ( q → ( p → r )). Proof. Prove instead p → ( q → r ) , q, p, ¬ r ‘ ⊥ . But this is exactly as above! 4. (5 MARKS) Prove that A ∨ A ∨ A ‘ B → A Proof. (1) A ∨ A ∨ A h hyp i (2) A ∨ A ∨ A ∨ A h (1) + X ‘ X ∨ Y i (3) A ∨ A h (2) + X ∨ X ‘ X i (4) A h (3) + X ∨ X ‘ X i (5) B → A h (4) + X ‘ Y ∨ X i Or just say: “Since obviously A ∨ A ∨ A | = taut B → A (any v that makes A ∨ A ∨ A t makes A t , and hence makes B → A t ), we have A...
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This note was uploaded on 05/09/2010 for the course LAPS CSE 1090 taught by Professor Peter during the Winter '10 term at York University.

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asg3-sol - MATH1090 Problem Set 3 —Solutions November...

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