hw1_solutions

# hw1_solutions - Problem Set 1 Solutions Note In the truth...

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Problem Set 1 Solutions Note: In the truth tables, I’m using 0 for F and 1 for T . Problem 1 1. R Q P 2. ( ¬ R ∨¬ Q ) ∨¬ P 3. ( ¬ R ∨¬ Q ) →¬ P None of the three statements are equivalent. P Q R ¬ P ¬ Q ¬ R R Q ¬ R ∨¬ Q R Q P ( ¬ R ∨¬ Q ) ∨¬ P ¬ R ∨¬ Q →¬ P 0 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 0 1 1 1 1 0 1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 0 1 0 0 1 1 1 0 0 0 1 1 0 1 1 1 0 1 0 1 0 1 0 0 1 1 1 0 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0 0 0 1 0 1 0 1 Problem 2 1. P Q is equivalent to ( P Q ) ( P Q ). P Q P Q P Q ( P Q ) ( P Q ) ( P Q ) ( P Q ) 0 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 0 0 1 2. P Q is equivalent to ( P P ) ( Q Q ). P Q P Q P P P P Q Q Q Q ( P P ) ( Q Q ) ( P P ) ( Q Q ) 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 1 1 0 1 0 0 1 3. ¬ P is equivalent to P P . P ¬ P P P P P 0 1 0 1 1 0 1 0 1

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4. T is equivalent to ( P P ) P . P P P P P ( P P ) P ( P P ) P T 0 0 1 0 1 1 1 1 0 0 1 1 5. F is equivalent to (( P P ) P ) (( P P ) P ). Note: The columns for P P , P P , ( P P ) P are omitted here. They are the same as in part 4 above. P ( P P ) P (( P P ) P ) (( P P ) P ) (( P P ) P ) (( P P ) P ) F 0 1 1 0 0 1 1 1 0 0 Problem 3 P Q R P Q Q R ( P
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hw1_solutions - Problem Set 1 Solutions Note In the truth...

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