a2notes - Math 486 Homework 2 Notes Paul Shafer...

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Math 486 Homework 2 Notes Paul Shafer pshafer@math.cornell.edu February 12, 2008 Problem I.3.3 We show that the proposition α has a CNF by using the fact that ¬ α has a DNF. Let ¬ α ( α 1 , 1 ∧ ··· ∧ α 1 ,n 1 ) ∨ ··· ∨ ( α k, 1 ∧ ··· ∧ α 1 ,n k ) be the DNF of ¬ α , where each α i,j is a literal. Then α ≡ ¬¬ α ≡ ¬ [( α 1 , 1 ∧ ··· ∧ α 1 ,n 1 ) ∨ ··· ∨ ( α k, 1 ∧ ··· ∧ α 1 ,n k )] ≡ ¬ ( α 1 , 1 ∧ ··· ∧ α 1 ,n 1 ) ∧ ··· ∧ ¬ ( α k, 1 ∧ ··· ∧ α 1 ,n k ) ( ¬ α 1 , 1 ∨ ··· ∨ ¬ α 1 ,n 1 ) ∧ ··· ∧ ( ¬ α k, 1 ∨ ··· ∨ ¬ α 1 ,n k ) This is almost the CNF of α . What some students failed to notice is that if α i,j is the literal ¬ P , then ¬ α i,j = ¬¬ P is not yet a literal. Thus we need to replace each ¬¬ P with P to put the formula in CNF. Problem I.3.5(iv) We show Cn(Σ) = Cn(Cn(Σ)). By I.3.5(ii) we know Σ Cn(Σ). Applying I.3.5(i) then gives Cn(Σ)
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This homework help was uploaded on 02/23/2008 for the course MATH 4860 taught by Professor Dorais during the Spring '08 term at Cornell University (Engineering School).

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a2notes - Math 486 Homework 2 Notes Paul Shafer...

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