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practicefinalsol

# practicefinalsol - CSE547 PRACTICE FINAL SOLUTIONS QUESTION...

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Unformatted text preview: CSE547 PRACTICE FINAL SOLUTIONS QUESTION 1 1. For the sentence If it is not true that: 2 + a = a + 3 and today is Monday, then: 2 + a 6 = a + 3 or today is not Monday. write its corresponding formula A . Solution The formula A is: ( ¬ ( a ∩ b ) ⇒ ( ¬ a ∪ ¬ b )) 2. Define a formal language to which the formula A belongs. The language is: L ¬ , ∩ ,cup, ⇒ QUESTION 2 Write the formula A from Question 1 as a formula of the language L {¬ , ∪} , i.e. as a formula B of L {¬ , ∪} , such that A ≡ B . Write down all logical equivalences you need while solving this problem. Solution ( ¬ ( a ∩ b ) ⇒ ( ¬ a ∪ ¬ b )) ≡ defimpl ( ¬¬ ( a ∩ b ) ∪ ( ¬ a ∪ ¬ b )) ≡ deMorgan ( ¬ ( ¬ a ∪ ¬ b ) ∪ ( ¬ a ∪ ¬ b )) QUESTION 3 H is the following proof system: S = ( L {⇒ , ¬} , A 1 ,A 2 ,A 3 , MP ) A1 ( A ⇒ ( B ⇒ A )) , A2 (( A ⇒ ( B ⇒ C )) ⇒ (( A ⇒ B ) ⇒ ( A ⇒ C ))) , A3 (( ¬ B ⇒ ¬ A ) ⇒ (( ¬ B ⇒ A ) ⇒ B ))) MP Rule of inference: ( MP ) A ; ( A ⇒ B ) B We know that S is SOUND and COMPLETE under classical semantics. Show whether S is sound/not sound under M semantics defined below. M Negation: ¬ F = T, ¬ ⊥ = ⊥ , ¬ T = F , M Conjunction: for any a,b ∈ { F, ⊥ ,T } , a ∩ b = min { a,b } , M Disjunction: for any a,b ∈ { F, ⊥ ,T } , a ∪ b = max { a,b } , 1 M Implication: for any a,b ∈ { F, ⊥ ,T } , a ⇒ b = ¬ a ∪ b . Solution S is NOT SOUND. Axiom A 1 is not a M-tautology. If A = ⊥ and B = ⊥ , then ( ⊥⇒ ( ⊥⇒⊥ )) = ⊥ . QUESTION 4 Here are consecutive steps B 1 ,...,B 5 in a proof of ( B ⇒ ¬¬ B ) in H 2 . The comments included are incomplete. Complete the comments by writing all details for each step of the proof. You have to write down the proper substitutions and formulas used at each step of the proof. B 1 = (( ¬¬¬ B ⇒ ¬ B ) ⇒ (( ¬¬¬ B ⇒ B ) ⇒ ¬¬ B )) Axiom A 3 B 2 = ( ¬¬¬ B ⇒ ¬ B ) Already proved fact: ‘ H 2 ( ¬¬ B ⇒ B ) B 3 = (( ¬¬¬ B ⇒ B ) ⇒ ¬¬ B ) (MP) B 4 = ( B ⇒ ( ¬¬¬ B ⇒ B )) Axiom A 1 B 5 = ( B ⇒ ¬¬ B ) Already proved fact: ( A ⇒ B ) , ( B ⇒ C ) ‘ H 2 ( A ⇒ C ) item[Solution] The completed comments are as follows. B 1 = (( ¬¬¬ B ⇒ ¬ B ) ⇒ (( ¬¬¬ B ⇒ B ) ⇒ ¬¬ B )) Axiom A 3 for A = B,B = ¬¬ B B 2 = ( ¬¬¬ B ⇒ ¬ B ) Already proved fact: ‘ H 2 ( ¬¬ B ⇒ B ) for B = ¬ B B 3 = (( ¬¬¬ B ⇒ B ) ⇒ ¬¬ B ) B 1 ,B 2 and MP, i.e. ( ¬¬¬ B ⇒ ¬ B );(( ¬¬¬ B ⇒ ¬ B ) ⇒ (( ¬¬¬ B ⇒ B ) ⇒ ¬¬ B )) (( ¬¬¬ B ⇒ B ) ⇒ ¬¬ B ) B 4 = ( B ⇒ ( ¬¬¬ B ⇒ B )) Axiom A 1 for A = B,B = ¬¬¬ B B 5 = ( B ⇒ ¬¬ B ) B 3 ,B 4 and already proved fact: ( A ⇒ B ) , ( B ⇒ C ) ‘ H 2 ( A ⇒ C ) for A = B,B = ( ¬¬¬ B ⇒ B ) ,C = ¬¬...
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practicefinalsol - CSE547 PRACTICE FINAL SOLUTIONS QUESTION...

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