Determine the truth value of $$(\forall x)(\exists y)(x=y\to x>y)$$ over the integers.
My work:
$(\forall x)(\exists y)(x=y\to x>y)$ means "For all $x$, there exists a $y$ where $x=y\to x>y$".
Since $x=y$, we can replace the $y$ with the $x$ and vice-versa. We can get $$x>x$$ once we replaced.
This statement is always false, no matter what $x$ is. So in this case $x=y$ is true, but $x>y$ is not true, or false. So, using the truth table below, $$\begin{array}{c|c|c}x=y&x>y&x=y\to x>y\\\hline\mathrm{T}&\mathrm{T}&\mathrm{T}\\\color{red}{\mathrm{T}}&\color{red}{\mathrm{F}}&\color{red}{\mathrm{F}}\\\mathrm{F}&\mathrm{T}&\mathrm{T}\\\mathrm{F}&\mathrm{F}&\mathrm{T}\end{array}$$ we can say that this statement is false.
Best Answer
Hint: $\forall x~\color{red}\exists y~(P(x,y)\to Q(x,y))$ is true exactly when: every $x$ has some $y$ where $Q(x,y)$ is true or $P(x,y)$ is false.
No. What the truth table is telling you is that the implication is true in three cases.
Case 1: $x=y$ and $x>y$. Well, for any $x$ that cannot happen. However we must check the other two cases.
Case 2: $\lnot(x=y)$ and $x>y$. Well, for any $x$ there is some $y$ where both of these are true.
Case 3: $\lnot (x=y)$ and $\lnot(x>y)$. Well, for any $x$ there is some $y$ where both of these are true.
Since for any $x$ there is a $y$ where $\lnot (x=y)$, then cases $2$ and $3$ may be satisfied for any $x$. Thus we confirm that for any $x$ there is some $y$ where the implication holds.