Question

Which of the following Boolean expressions is not a tautology?

Options

• `(p \implies q) ∨ (∼q \implies p)`

• `(q \implies p) ∨ (∼q \implies p)`

• `(p \implies ∼q) ∨ (∼q \implies p)`

• `(∼p \implies q) ∨ (∼q \implies p)`

Answer 1

`bb((∼p \implies q) ∨ (∼q \implies p))`

Explanation:

(1) (p `rightarrow` q) ∨ (∼q `rightarrow` p) = (∼p ∨ q) ∨ (q ∨ p)

= (∼p ∨ p) ∨ q

= t ∨ q

= t

It is tautology.

(2) (q `rightarrow` p) ∨ (∼q `rightarrow` p) = (∼q ∨ p) ∨ (q ∨ p)

= (∼q ∨ q) ∨ p

= t ∨ t

= t

It is tautology.

(3) (p `rightarrow` ∼q) ∨ (∼q `rightarrow` p) = (∼p ∨ ∼q) ∨ (q ∨ p)

= (∼p ∨ q) ∨ (∼q ∨ q)

= t ∨ t

= t

It is tautology.

(4) (∼q `rightarrow` q) ∨ (∼q `rightarrow` p) = (p ∨ q) ∨ (q ∨ p)

= (p ∨ p) ∨ (q ∨ p)

= p ∨ q

Which is not a tautology.