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Write the negation of p ↔ q.

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Question

Write the negation of p ↔ q.

Sum
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Solution

∼(p ↔ q)

Using the equivalence

p ↔ q ≡ (p → q) ∧ (q → p)

we get

∼ [(p → q) ∧ (q → p)]

Replace implications:

= ∼[(∼ p ∨ q) ∧ (∼ q ∨ p)]

Apply De Morgan’s law:

= ∼(∼ p ∨ q) ∨ ∼ (∼ q ∨ p)

Again using De Morgan’s law:

= (p ∧ ∼ q) ∨ (q ∧ ∼ p)

So,

∼ (p ↔ q) ≡ (p ∧ ∼ q) ∨ (q ∧ ∼ p)​ which is the XOR form (“p and q are different”).

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