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Let $P(x)$ and $Q(x)$ be arbitrary predicates. Which of the following statements is always **TRUE**?

- $\left(\left(\forall x \left(P\left(x\right) \vee Q\left(x\right)\right)\right)\right) \implies \left(\left(\forall x P\left(x\right)\right) \vee \left(\forall xQ\left(x\right)\right)\right)$
- $\left(\forall x \left(P\left(x\right) \implies Q\left(x\right)\right)\right) \implies \left(\left(\forall x P\left(x\right)\right) \implies \left(\forall xQ\left(x\right)\right)\right)$
- $\left(\forall x\left(P\left(x\right) \right) \implies \forall x \left( Q\left(x\right)\right)\right) \implies \left(\forall x \left( P\left(x\right) \implies Q\left(x\right)\right)\right)$
- $\left(\forall x \left( P\left(x\right)\right) \Leftrightarrow \left(\forall x \left( Q\left(x\right)\right)\right) \right) \implies \left(\forall x \left (P\left(x\right) \Leftrightarrow Q\left(x\right)\right)\right)$

66 votes

Best answer

**Procedure to be followed**: For each option assume LHS to be TRUE and try to make RHS be False by selecting some values which makes LHS true in every condition.

(one can also start from RHS assuming it to be false and trying to make LHS to be true)

A. $\mathbf{[ (\forall x (P(x)} ∨ \mathbf{Q(x))) ]} ⟹ [ (∀ \mathbf{x P(x))} ∨ \mathbf{(∀xQ(x))} ]$

Let us assume the domain of $x$ such that, for the first half of the values $P(x)$ is True & $Q(x)$ is False while for the other half, $Q(x)$ is True and $P(x)$ is False.

LHS: Since for LHS to be true either $P(x)$ or $Q(x)$ should be true. Hence our assumption of the domain will make LHS be TRUE.

RHS: for $(∀\mathbf{x P(x))}$** **or** **$(∀ \mathbf{x Q(x))}$** **to be true, atleast one of the $P(x)$ or $Q(x)$ must be true for all values of $x$, which is not possible as per assumption we have made.Thus, RHS becomes FALSE.

Thus, $T → F$ makes **statement A False.**

C. $[ ∀\mathbf{x(P(x))} ⟹ ∀\mathbf{x(Q(x))} ] ⟹ [ ∀\mathbf{x(P(x)⟹Q(x))} ]$

Let us assume some values of $P(x)$ and $Q(x)$ as follows :

$$\begin{array}{l|l|l} \text{$x$} & \text{$P(x)$} & \text{ $Q(x)$} \\ \hline \text{$x_1$} & \text{$F$} & \text{$T$} \\ \text{$x_2$} & \text{$T$} &\text{$F$} \end{array}$$

LHS: for assumed domain $∀ \mathbf{ x(P(x)) }$ will becomes False and $∀\mathbf{x(Q(x))}$ will also be false. Since $F → F$ is True, LHS becomes TRUE.

RHS: for $x_1$ $\mathbf{[(P(x)⟹Q(x))}$ is True and for $x_2$ $\mathbf{(P(x)⟹Q(x))}$ is False. As a whole** **$∀\mathbf{x(P(x)⟹Q(x))}$ becomes False, thus RHS becomes FALSE**.**

Thus, $T→F$ makes **statement C as False**.

D. $[ ∀\mathbf{x(P(x))} ⇔ \mathbf{(∀x(Q(x))) ]} ⟹ [ ∀\mathbf{x(P(x) ⇔ Q(x)) ]}$

if we assume same domain as in above option C,then observations are as following :

LHS: $∀\mathbf{x(P(x))}$ becomes false as $x_1$ is false. Also $(∀\mathbf{x(Q(x)))}$ becomes false as $x_2$ is false.Thus $F ↔ F$ implies LHS is TRUE.

RHS: $\mathbf{(P(x) ⇔ Q(x))}$** **will be false for both $x_1$ and $x_2$. Hence $∀\mathbf{x(P(x) ⇔ Q(x))}$** **becomes False which makes RHS to be FALSE.

Thus $T→F$ makes **statement D as False.**

B. $[ ∀\mathbf{x(P(x) ⟹ Q(x)) ] ⟹ [ (∀xP(x)) ⟹ (∀xQ(x)) ]}$

As we are assuming LHS to be TRUE then we'll not make any selection in which $P(x)$ is True and $Q(x)$ is false as it will make our assumption false.Thus values can be like this:

$$\begin{array}{l|l|l}\text{$x$} & \text{$P(x)$} & \text{ $Q(x)$} \\\hline \text{$x_1$} & \text{$T$} & \text{$T$} \\ \text{$x_2$} & \text{$F$} &\text{$T$} \\ \text{$x_3$} & \text{$F$} &\text{$F$} \end{array}$$

LHS: $\mathbf{(P(x) ⟹ Q(x))}$ becomes TRUE for each value and thus** **$∀\mathbf{x(P(x) ⟹ Q(x))}$ become TRUE.

RHS: $\mathbf{(∀xP(x)) \text{ and } (∀xQ(x))}$ both becomes false for assumed values which implies $F→F$ and thus makes RHS to be TRUE.

$T→T$ makes **statement B to be TRUE. Thus Answer is B.**

63 votes

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