Set theory?

Obs: $|x|$ means the set that contains $x$ . Well, as we know from the problem:

$Y^{\phi} \subset P(\phi \times Y) = P(\phi) = |\phi| \therefore Y^{\phi} \subset |\phi|$

So either $Y^{\phi} = \phi$ or else $Y^{\phi} = |\phi|$ . But $f = \phi$ satisfies $f: \phi \rightarrow Y$ , therefore $Y^{\phi}$ has at least one element. But $Y^{\phi} = |\phi|$ is the only set that satisfies this condition, therefore $N = 1$ . But now observe that

$\phi^{X} \subset P(X \times \phi) = P(\phi) = |\phi| \therefore \phi^{X} \subset |\phi|$

So either $\phi^{X} = |\phi|$ or else $\phi^{X} = \phi$ . But if $X \neq \phi$ , then for some $x \in X$ , there is no $y \in Y$ (because $Y = \phi$ . Therefore there does not exist $f$ such that

$f: X \neq \phi \rightarrow \phi$

And then $\phi^{X} = \phi$ which means that $M = 0$ . We want $N + M = 1 + 0 = 1$ .