DISCRETE MATH PROBLEM: Simplify ~(r ∨ s) ∨ (~s ∧ r). Indicate, at each step, which logical equivalence law is used. EXAMPLE BELOW: Simplify p∧ ∼ (p∧ ∼ q). Indicate, at each step, which logical equivalence law is used. The expression can be simplified as follows: 1. p∧ ∼ (p∧ ∼ q), Given Expression 2. p ∧ (∼ p∨ ∼ (∼ q), De Morgan’s Law 3. p ∧ (∼ p ∨ q), Double Negation Now, use the distributive law to simplify further: 1. (p∧ ∼ p) ∨ (p ∧ q), Distributive Law 2. c ∨ (p ∧ q), Negation Law 3. (p ∧ q), Identity Law Therefore, (p ∧ q) is the simplified expression.
DISCRETE MATH PROBLEM: Simplify ~(r ∨ s) ∨ (~s ∧ r). Indicate, at each step, which logical equivalence law is used. EXAMPLE BELOW: Simplify p∧ ∼ (p∧ ∼ q). Indicate, at each step, which logical equivalence law is used. The expression can be simplified as follows: 1. p∧ ∼ (p∧ ∼ q), Given Expression 2. p ∧ (∼ p∨ ∼ (∼ q), De Morgan’s Law 3. p ∧ (∼ p ∨ q), Double Negation Now, use the distributive law to simplify further: 1. (p∧ ∼ p) ∨ (p ∧ q), Distributive Law 2. c ∨ (p ∧ q), Negation Law 3. (p ∧ q), Identity Law Therefore, (p ∧ q) is the simplified expression.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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DISCRETE MATH PROBLEM:
Simplify ~(r ∨ s) ∨ (~s ∧ r). Indicate, at each step, which logical equivalence law is used.
EXAMPLE BELOW:
Simplify p∧ ∼ (p∧ ∼ q). Indicate, at each step, which logical equivalence law is used.
The expression can be simplified as follows:
1. p∧ ∼ (p∧ ∼ q), Given Expression
2. p ∧ (∼ p∨ ∼ (∼ q), De Morgan’s Law
3. p ∧ (∼ p ∨ q), Double Negation
Now, use the distributive law to simplify further:
1. (p∧ ∼ p) ∨ (p ∧ q), Distributive Law
2. c ∨ (p ∧ q), Negation Law
3. (p ∧ q), Identity Law Therefore, (p ∧ q) is the simplified expression.
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