# Determine the truth value of $(\forall x)(\exists y)(x=y\to x>y)$ over the integers

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.

(Image that replaced text).

• What you wrote in that solution explains why $x=y\to x>y$ is not always true. So that shows that $\forall x\forall y(x=y\to x>y)$ is false, but it doesn't say anything about the given problem. Jul 13 '19 at 23:31
• What do you mean by that? Jul 13 '19 at 23:32
• I mean your solution is wrong. Jul 13 '19 at 23:33
• You don't. It's true. Jul 13 '19 at 23:34
• What was wrong about the first solution was you sort of missed the point to the $\forall x\exists y$ part (the "quantifiers"). Here's an outline of the solution that deals properly with the quantifiers, with blanks you should be able to fill in: "Given $x$, let $y=\dots$. Then ..., so $x=y\to x>y$ is true." Jul 13 '19 at 23:42

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.

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.

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.

If $$P(x)$$ is false for some $$x$$, then $$P(x) \to Q(x)$$ is true no matter what the value of $$Q(x)$$ is so if you find an $$y$$ where $$y \neq x$$(e.g $$x-1$$), the statement would always hold.