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

True

To prove that $\forall x \forall y P(x,y)$ is true
Domain of discourse is $X\times Y$
You must show that $P(x,y)$ is true for all values of $x \in X$ and $y \in y$ . — ระวังพวกกำลัง 2
One technique is to argue that $P(x,y)$ is true using the symbols $x$ and $y$ to stand for arbitrary elements in $X$ and $Y$ .

To prove that $\forall x \forall y P(x,y)$ is false
Domain of discourse is $X\times Y$
Find one value of $x \in X$ $x \in X$ and one value of $y \in Y$ $y \in Y$
- (two values suffice—one for x and one for y) that make $P(x,y)$ false.

To prove that $\forall x \exists y P(x,y)$ is true
Domain of discourse is $X\times Y$
you must show that for all $x \in X$ , there is at least one $y \in Y$ such that $P(x,y)$ is true.
One technique is to let x stand for an arbitrary element in X and then ﬁnd a value for $y \in Y$ (one value suffices!) that makes $P(x,y)$ true.

To prove that $\forall x \exists y P(x,y)$ is false
Domain of discourse is $X\times Y$
You must show that for at least one $x \in X$ , $P(x,y)$ is false for every $y \in Y$ .
One technique is to find a value of $x \in X$ $x \in X$ (again one value suffices!) that has the property that $P(x,y)$ $P (x, y)$ is false for every $y \in Y$ $y \in Y$ .
- Having chosen a value for $x$ , let $y$ stand for an arbitrary element of $Y$ and show that $P(x,y)$ is always false.

To prove that $\exists x \forall y P(x,y)$ is true
Domain of discourse is $X\times Y$
You must show that for at least one $x \in X$ , $P(x,y)$ is true for every $y \in Y$ .
One technique is to ﬁnd a value of $x \in X$ (again one value suffices!) that has the property that $P(x,y)$ is true for every $y \in Y$ .
Having chosen a value for $x$ , let $y$ stand for an arbitrary element of $Y$ and show that $P(x,y)$ is always true.

To prove that $\exists x \forall y P(x,y)$ is false
Domain of discourse is $X\times Y$
You must show that for all $x \in X$ , there is at least one $y \in Y$ such that $P(x,y)$ is false.
One technique is to let $x$ stand for an arbitrary element in $X$ and then ﬁnd a value for $y \in Y$ (one value suffices!) that makes $P(x,y)$ false.

To prove that $\exists x \exists y P(x,y)$ is true
Domain of discourse is $X\times Y$
Find one value of $x \in X$ and one value of $y \in Y$ (two values suffice—one for x and one for y) that make $P(x,y)$ true.

To prove that $\exists x \exists y P(x,y)$ is false
Domain of discourse is $X\times Y$
You must show that $P(x,y)$ is false for all values of $x \in X$ and $y \in Y$ .
One technique is to argue that $P(x,y)$ is false using the symbols $x$ and $y$ to stand for arbitrary elements in X and Y.