Solution

a) $Q(1, 1) \Bi 1+1=1-1$. $Q(1, 1)$ is false.
b) $Q(2, 0) \Bi 2+0=2-0$. $Q(2, 0)$ is true.
c) $\forall y Q(1, y)$ denotes the proposition "For every integer y, $1+y=1-y$ holds". When $y=1$, $1+y=1-y$ is false, therefore $\forall y Q(1, y)$ is false.
d) $\exists x Q(x, 2)$ denotes the proposition "There is an integer x, such that $x+2=x-2$". Such $x$ can not be found, $\exists x Q(x, 2)$ is false.
e) $\exists x \exists yQ(x, y)$ denotes the proposition "There is a pair $x, y$ such that $Q(x, y)$". The pair $x=1, y=0$ satisfies $Q(x, y)$, $\exists x \exists yQ(x, y)$ is true.
f) $\forall x\exists yQ(x, y)$ denotes the proposition "For every integer $x$ there is an integer $y$ such that $Q(x, y)$". When $y=0$, $Q(x, y)$ is true for every integer $x$, $\forall x\exists yQ(x, y)$ is true.
g) $\exists y \forall x Q(x, y)$ denots the proposition "There is an integer $y$, such that for every integer $x$, $x+y=x-y$ holds". When $y=0$, $x+y=x-y$ holds for every $x$, $\exists y \forall x Q(x, y)$ is true.
h) $\forall y\exists xQ(x, y)$ denotes the proposition "For every integer $y$ there is an integer $x$ such that $Q(x, y)$". When $y=1$, we could not find any $x$ to satisfy $Q(x, y)$, $\forall y\exists xQ(x, y)$ is false.

Solution

e) The quantification $\exists n\exists m(n^2 + m^2 = 5)$ denotes the proposition "There is a pair $m, n$ such that $n^2 + m^2 = 5$". The pair $m=1, n=2$ satisfies $n^2 + m^2 = 5$. Hence the statement is true.
g) The quantification $\exists n\exists m(n + m = 4 \land n - m = 1)$ denotes the proposition "There is a pair $m, n$ such that $n + m = 4 \land n - m = 1$". We could not find any integer pair $m,n$ to satisfy $n + m = 4 \land n - m = 1$, so the statement is false.
h) The quantification $\exists n\exists m(n + m = 4 \land n - m = 2)$ denotes the proposition "There is a pair $m, n$ such that $n + m = 4 \land n - m = 2$". The pair $m=1,n=3$ satisfies $n + m = 4 \land n - m = 2$. Hence, the statement is true.

Solution

$\begin{aligned} \neg (\forall x\exists yP (x, y) \lor \forall x\exists yQ(x, y)) &\equiv \neg \forall x\exists yP (x, y) \land \neg \forall x\exists yQ(x, y) \\ &\equiv \exists x \neg \exists yP (x, y) \land \exists x \neg \exists yQ(x, y) \\ &\equiv \exists x \forall y \neg P (x, y) \land \exists x \forall y \neg Q(x, y) \\ \end{aligned}$

Solution

$\begin{aligned} \neg \forall x\exists y(P (x, y) \to Q(x, y)) &\equiv \exists x \neg \exists y(P (x, y) \to Q(x, y)) \\ &\equiv \exists x \forall y \neg(P (x, y) \to Q(x, y)) \\ &\equiv \exists x \forall y (P (x, y) \land \neg Q(x, y)) \end{aligned}$

Solution

The quantification $\exists x\forall y(x \leqslant y^2)$ denotes the proposition "There is a $x$ such that for every $y$, $x \leqslant y^2$".
a) The statement is false when $x,y$ are positive numbers. Because $y^2 \geqslant 0$, we could not find any positive number $x$ that $x \leqslant y^2$.
b) The statement is true when $x,y$ are integers. When $x=0$, the $x \leqslant y^2$ is true for every $y$.
c) The statement is true when $x,y$ are nonzero real numbers. When $x=-1$, the $x \leqslant y^2$ is true for every $y$.