$7$ is prime

$4$ is prime

$\varnothing \subseteq \{1\}$

$n$ is prime

No, union takes only what is unshared

It cannot be decided

Yes, and lying in both sets is no obstacle

No, because $2$ lies in both

False, since $S$ is empty

It has no truth value

True, there is nothing to violate it

False, since $S$ has no negative elements

Yes, both directions hold here

The converse has no truth value

No, since $x = -2$ also squares to $4$

Yes, converses inherit truth

$\forall x\, \exists y\, \neg P(x,y)$

$\exists x\, \exists y\, \neg P(x,y)$

$\forall x\, \forall y\, \neg P(x,y)$

$\exists x\, \forall y\, \neg P(x,y)$

$\{1,2\}$

$\{1\}$

$\{1,2,3\}$

$\{5,7\}$

Assume $\neg Q$ and derive $\neg P$

Assume $P$ and derive $Q$

Assume $Q$ and derive $P$

Assume $P$ and $\neg Q$

$\forall x\, \exists y\, (y > x)$

$\exists x\, (x > 5)$

$\forall x\, (x \ge 0 \lor x < 0)$

$\exists y\, \forall x\, (y > x)$

$\exists x\,\exists y\,\big(\neg Q(x,y)\lor\neg R(y)\big)$

$\exists x\,\forall y\,\big(\neg Q(x,y)\lor\neg R(y)\big)$

$\exists x\,\forall y\,\big(\neg Q(x,y)\land\neg R(y)\big)$

$\forall x\,\exists y\,\big(\neg Q(x,y)\lor\neg R(y)\big)$

True

False

finite $\implies$ countable

countable $\implies$ finite

finite $\iff$ countable

not countable $\implies$ finite

$\forall a\,\exists b\,\big(f(a)=b\big)$

$\exists b\,\forall a\,\big(f(a)=b\big)$

$\forall b\,\exists a\,\big(f(a)=b\big)$

$\exists a\,\forall b\,\big(f(a)=b\big)$