Show that ( pimpliesq)^(qimplies~ p) equivalent~ p i) with truth table. ii) without truth table.

a) The statement "p implies q" means that if p is true, then q must also be true.

$\begin{matrix} p & q & \sim p & p \ implies\ q & q \ implies \sim p & p \ implies\ q \cap q \ implies \sim p\\ T & T & F & T & F & F \\ T & F & F & F & T & F \\ F & T & T & T & T & T \\ F & F & T & T & T & T \end{matrix}$

From above truth table, it is clear that 3rd column and last column are same, hence

( pimpliesq)^(qimplies~ p) equivalent~ p

b) p implies q is false when p is true and q is false, so in that case ( pimpliesq)^(qimplies~ p) will be false. Another case where ( pimpliesq)^(qimplies~ p) is false when (qimplies~ p) is false which is true if and only if q is true and $\sim$ p is fasle (p is true).

Hence, ( pimpliesq)^(qimplies~ p) equivalent~ p is false iff p is true case. Thus

( pimpliesq)^(qimplies~ p) equivalent ~ p.