Here, domain is set of integers, So Elements

$$ x, y E\{\ldots . .-3,-2,-1,0,1,2, \ldots \ldots\} $$

Expression $E=\forall x\left\{P(x) \Rightarrow \exists_y Q(x, y)\right\}$

Which says if x is P then 'always exists a y such such that $Q(x, y)$.

Now, checking options.

For option (A) :

Is $\frac{\exists[P(x)] \wedge \forall_y \theta(x, y)}{\text { LHS }} \rightarrow \frac{E}{\text { RHS }}$ True ?

Here, for LHS to be true. Say there exists an $x=(6)$ for which $p(x)=p(c)=$ True and for all $y \theta(6, y)$ is True.

Now, RHS : $E=\forall x<\frac{p(x)}{A} \Rightarrow \frac{\exists y 2(x, y)}{B}$

$A=$ True for say $x=7$, so $p(7)=$ True

For B, say there doesn't exist any y such that $Q(7, y)=$ True. Hence $A \Rightarrow B$

$T \Rightarrow F$ is false.

So, its case of True $\Rightarrow$ False as LHS is True and A

RHS is False

Therefore it becomes case of True $\rightarrow$ False and eventually its False:

For option (B) :

$$ \frac{\forall x \forall y Q(x, y)}{L H S} \rightarrow \frac{E}{R H S} \text { True? } $$

Here, LHS is True for all values of x and y .

Now Ans: $E=\forall x \frac{[P(x)]}{A} \Rightarrow \frac{\exists y Q(x, y)]}{B}$

Now, since,

$$ \begin{aligned} & \forall x \forall y Q(x, y)=\text { True. } \\\\ & \exists y Q(x, y)=\text { True too. } \end{aligned} $$

So, For $A \Rightarrow B$

$\mathrm{A} \Rightarrow$ True is always True.

As RHS is True. Its case of

True $\rightarrow$ True which is true.

For option (C) :

$$ \frac{\exists y \forall x[P(x) \Rightarrow Q(x, y)]}{\text { LHS }} \rightarrow \frac{E}{R H S} \text { True? } $$

For LHS to be true, there exists some y say $\mathrm{y}=$ 2 for which for all x which are satisfying property p implies property $\mathrm{Q}(\mathrm{x}, \mathrm{y})$

Now, RHS : $=E=\forall x\left[\frac{P(x)}{A} \Rightarrow \frac{\exists y Q(x, y)}{B}\right]$

$B=\exists y Q(x, y)$ is always, true as there exists at least one $y$ (we assumed $\mathrm{y}=2$ ) such that $\forall x Q(x, 2)$ is True. So B is true

The case becomes $A \Rightarrow B$

$A \Rightarrow$ True which is true

So, L.H.S $\rightarrow$ R.H.S. is true.

For Option (D) :

$\frac{\exists \times[P(x) \wedge \exists y 2(x, y)]}{\text { LHS }} \rightarrow \frac{E}{R H S}$ True?

For LHS to be true, assume $x=6$ for which properly $P(6)$ is true and there exists a $y$ assume $y=2$ such that $Q(6,2)$ is true.

Now, RHS $=E=V x\left[\frac{P(x)}{A} \Rightarrow \frac{\exists y Q(x, y)}{B}\right]$

Say $x=3$ and $P(x)=$ True

But there exists no y for $Q(3, y)$ to be true.

Hence, it becomes $A \Rightarrow B$

True $\Rightarrow$ False so RHS is False

Also, it becomes case of True $\rightarrow$ False which is false.

Hence, the correct options are (B) and (C).