Asked by Joy
Let p and q be propositions. Use Logical equivalences to show that
(p ^ (~ (~ p V q))) V (p ^ q) = p
(p ^ (~ (~ p V q))) V (p ^ q) = p
Answers
Answered by
MathMate
The idea is to reduce the nesting of parentheses which will then give possibilities of simplification.
Remember to name the rule of simplification as justification.
Start with the left-hand side:
(p ^ (~ (~ p V q))) V (p ^ q)
de Morgan's law:
= (p ^ ( p ^ ~q)) V (p ^ q)
associative properties of ^:
= (p ^ p ^ ~q ) V (p ^ q)
idempotent law (p^p)=p
= ( p ^ ~q ) V (p^q)
distributive law:
= [p V (p^q)] ^ [~q V (p^q)]
distributive law:
= [(pVp)^(pVq)] ^ [(~qVp)^(~qVq)]
associative law:
= (pVp) ^ (pVq) ^ (pV~q) ^ (~qVq)
distributive law:
= (pVp) ^ pV(q^~q) ^ (~qVq)
complement laws:
= (pVp) ^ (pVF) ^ T
identity laws:
= (pVp) ^ p
idempotent law:
= p^p
idempotent law:
= p
QED
Remember to name the rule of simplification as justification.
Start with the left-hand side:
(p ^ (~ (~ p V q))) V (p ^ q)
de Morgan's law:
= (p ^ ( p ^ ~q)) V (p ^ q)
associative properties of ^:
= (p ^ p ^ ~q ) V (p ^ q)
idempotent law (p^p)=p
= ( p ^ ~q ) V (p^q)
distributive law:
= [p V (p^q)] ^ [~q V (p^q)]
distributive law:
= [(pVp)^(pVq)] ^ [(~qVp)^(~qVq)]
associative law:
= (pVp) ^ (pVq) ^ (pV~q) ^ (~qVq)
distributive law:
= (pVp) ^ pV(q^~q) ^ (~qVq)
complement laws:
= (pVp) ^ (pVF) ^ T
identity laws:
= (pVp) ^ p
idempotent law:
= p^p
idempotent law:
= p
QED
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