Question

Which one of the following predicate formulae is NOT logically valid?
Note that W is a predicate formula without any free occurrence of x.

• A. $\exists x(p\left(x\right)\wedge W)\equiv \exists xp\left(x\right)\wedge W$
• B. $\forall x(p\left(x\right)\to W)\equiv \forall xp\left(x\right)\to W$
• C. $\exists x(p\left(x\right)\to W)\equiv \forall xp\left(x\right)\to W$
• D. $\forall x(p\left(x\right)\vee W)\equiv \forall xp\left(x\right)\vee W$

Answer 1

The correct option is B $\forall x(p\left(x\right)\to W)\equiv \forall xp\left(x\right)\to W$

$\forall x\left(p\right(x\left)\right)\to W)\Rightarrow (\forall x\left(p\right(x\left)\right)\to$ W is true.
But $\forall x\left(p\right(x\left)\right)\to W\Rightarrow \forall x\left(p\right(x)\to W)$ is false.
Since if LHS is true then $(P_1\wedge P_2\wedge P_3...\wedge P_n)\to$ W will be true and RHS will be false since P${}_1\to$ W, itself will be false, since only P${}_1$ true cannot make W true (we need all the $P_1,P_2...,P_n$ to be true to make W true )
So LHS $\Rightarrow$ RHS true and RHS $\Rightarrow$ LHS is false so LHS $\ne$ RHS .
So option (b) is invalid.
Note that option (c) is true bacause by Boolean algebra,
$LHS=\exists x\left(p\right(x)\to W)\equiv \exists x(\sim p(x)\vee W)$
$\equiv \exists x(\sim p(x\left)\right)\vee W$
RHS = $\forall xp\left(x\right)\to W\equiv \sim \left(\forall xp\right(x\left)\right)\vee W$
$\equiv \exists x\sim p\left(x\right))\vee W$
So LHS = RHS.