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Unformatted text preview: valid. 1 2. (1.4 # 31, 15 points ) For the circuit corresponding to the Boolean expression ( ∼ P ∧ ∼ Q ) ∨ ( ∼ P ∧ Q ) ∨ ( P ∧ ∼ Q ) there is an equivalent circuit with at most two logic gates. Find such a circuit. Solution: Using the logical equivalences from Theorem 1.1.1, we can reduce the given Boolean expression as follows. ( ∼ P ∧ ∼ Q ) ∨ ( ∼ P ∧ Q ) ∨ ( P ∧ ∼ Q ) ≡ h ∼ P ∧ ( ∼ Q ∨ Q ) i ∨ ( P ∧ ∼ Q ) Distribution Law ≡ ( ∼ P ∧ T ) ∨ ( P ∧ ∼ Q ) Negation Law ≡ ∼ P ∨ ( P ∧ ∼ Q ) Identity Law ≡ ( ∼ P ∨ P ) ∧ ( ∼ P ∨ ∼ Q ) Distribution Law ≡ T ∧ ( ∼ P ∨ ∼ Q ) Negation Law ≡ ∼ P ∨ ∼ Q Identity Law ≡ ∼ ( P ∧ Q ) DeMorgan’s Law The corresponding circuit for the Boolean expression ∼ ( P ∧ Q ) is given by: P Q 2...
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This note was uploaded on 04/14/2008 for the course MATH 15A taught by Professor Briggs during the Fall '04 term at UCSD.
 Fall '04
 Briggs
 Math

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