Question

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

• A. $\left(\forall x(P\left(x\right)\vee Q\left(x\right))\right)\Rightarrow (\left(\forall xP\left(x\right)\right)\vee \left(\forall xQ\left(x\right)\right))$
• B. $\left(\forall x(P\left(x\right)\Rightarrow Q\left(x\right))\right)\Rightarrow (\left(\forall xP\left(x\right)\right)\Rightarrow \left(\forall xQ\left(x\right)\right))$
• C. $\left(\forall x(Px\Rightarrow \left(\forall xQ\left(x\right)\right))\right)\Rightarrow \left(\forall x(P\left(x\right)\Rightarrow Q\left(x\right))\right)$
• D. $((\forall x\left(P\left(x\right)\right)\Rightarrow \left(\forall xQ\left(x\right)\right))\Rightarrow \left(\forall x(P\left(x\right)\Rightarrow Q\left(x\right))\right)$

Answer 1

The correct option is B $\left(\forall x(P\left(x\right)\Rightarrow Q\left(x\right))\right)\Rightarrow (\left(\forall xP\left(x\right)\right)\Rightarrow \left(\forall xQ\left(x\right)\right))$

Consider choice (b)
$\left(\forall x\right(P\left(x\right)\Rightarrow Q\left(x\right)\left)\right)\Rightarrow \left(\right(\forall xP\left(x\right))\Rightarrow (\forall xQ\left(x\right)\left)\right)$
Let the LHS of this implication be true This means that
$P_1\to Q_1$
$P_2\to Q_2$
$⋮$
$P_n\to Q_n$
Now are need to check if the RHS is also true .
The RHS is $\left(\right(\forall xP\left(x\right))\Rightarrow (\forall xQ\left(x\right)\left)\right)$
To check this let us take the LHS of this as true
i.e. take $\forall xP\left(x\right)$ to be true. This means that $(P_1,P_2,...P_n)$ is taken to be true. Now $P_1$ along with $P_1\to Q_1$ will imply that $Q_1$ is true . Similarly $P_2$ along with $P_2\to Q_2$ will imply that $Q_2$ is true.
And so on....
Therefore ($Q_1$, $Q_2$,....$Q_n$) all true.
i.e. $\forall$xQ(x) is true. Therefore the statement (b) is valid predicate statement.