Question

Consider the following expressions:

(i) False
(ii) Q
(iii) True
(iv) $P\vee Q$
(v) $\rightharpoondown Q\vee P$

The number of expressions given above that are logically implied by $P\wedge (P\Rightarrow Q)$ is

• A. 4

Answer 1

The correct option is A 4

$p\wedge (p\Rightarrow q)=p(p{}^{\prime }+q)\equiv pq$
Take (i) false
$pq\Rightarrow$ false $\equiv pq\Rightarrow 0$
$\equiv \left(pq\right){}^{\prime }+0$
$\equiv p{}^{\prime }+q{}^{\prime }+0$
$\equiv$ not valid
Take (ii)
$pq\Rightarrow$ q $\equiv \left(pq\right){}^{\prime }+q$
$\equiv \left(pq\right){}^{\prime }+q$
$\equiv p{}^{\prime }+1+\equiv 1$
$\equiv$ valid
Take (iii)
$pq\Rightarrow$ true $\equiv pq\Rightarrow 1$
$\equiv \left(pq\right){}^{\prime }+1\equiv 1$
$\equiv$ valid
Take (iv)
$pq\Rightarrow$ p + q $\equiv \left(pq\right){}^{\prime }+p+q$
$\equiv p{}^{\prime }+q{}^{\prime }+p+q$
$\equiv 1+1$
$\equiv 1$
$\equiv$ valid

Take (v)
$pq\Rightarrow q{}^{\prime }+p\equiv \left(pq\right){}^{\prime }+p+q$
$\equiv p{}^{\prime }+q{}^{\prime }+p+q$
$\equiv 1$
$\equiv$ valid
So the number of expressions that are logically implies by $p\wedge (p\Rightarrow q)$ is 4