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The symbol ⊕ is used to denote exclusive or, so p ⊕ q ≡ (p v q) ^ ~ (p ^ q).
a) simplify the statement p ⊕ p.
b) Show that p ⊕ q ≡ ~ (p ↔ q) without using a truth table.
So how would this be done?

Sagot :

LammettHash

I'll use text forms of the logical symbols because copying/pasting on mobile is annoying.

⊕ = XOR

∨ = OR

∧ = AND

~ = NOT

↔ = IFF (meaning "if and only if")

→ = IMP (short for "implies")

and for logical equivalence I'll use a regular equal sign.

Then

p XOR q = (p OR q) AND NOT (p AND q)

(a) By definition of XOR, we have

p XOR p = (p OR p) AND NOT (p AND p)

= p AND NOT p

and this is tautologically FALSE.

(b) To rewrite XOR, first distribute the NOT over the AND using DeMorgan's law:

NOT (p AND q) = NOT p OR NOT q

Then we can interchange the AND and ORs using their respective distributive properties. That is,

p XOR q = (p OR q) AND (NOT p OR NOT q)

= (p AND (NOT p OR NOT q)) OR (q AND (NOT p OR NOT q))

= ((p AND NOT p) OR (p AND NOT q)) OR ((q AND NOT p) OR (q AND NOT q))

Both (p AND NOT p) and (q AND NOT q) are tautologically FALSE, so the truth value of either OR depends only on the other statements, so we have

p XOR q = (p AND NOT q) OR (q AND NOT p)

Recall the definition of IFF:

p IFF q = (p IMP q) AND (q IMP p)

Also recall that IMP is logically equivalent to

p IMP q = p OR NOT q

Then we have

p IFF q = (p OR NOT q) AND (q OR NOT p)

so that its negation is

NOT (p IFF q) = NOT ((p OR NOT q) AND (q OR NOT p))

= NOT (p OR NOT q) OR NOT (q OR NOT p)

= (NOT p AND q) OR (NOT q AND p)

The statements in bold match, so p XOR q is indeed equivalent to NOT (p IFF q).

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