Why is $(\forall x \in \mathbb{Z})(\exists y \in \mathbb{Z}) (2x + y = 3 \to x + 2y = 3)$ true? 
Is the statement true or false? $$(\forall x \in \mathbb{Z})(\exists y \in \mathbb{Z}) (2x + y = 3 \to x + 2y = 3)$$ The answer is that the entire statement is true, but I do not see why. 

The first part of the equation $2x + y = 3$ would be true since for every $x$ there is at least one that satisfies the equation. The right is false, since there for every $x$ there is not at least one $y$ that satisfies the equation. But this would make the conditional statement false since the left is true and the right is false? 
 A: No matter which $x$ you come with, I can choose an $y$ that makes both sides of the $\to$ false, and therefore makes the entire $\to$ true.
The meaning of formulas of the form $\exists x(\cdots \to \cdots)$ is not very intuitive; the $\to$ connective is really designed to live under a $\forall$ instead. In my experience, the best one can do for getting an intuitive handle on the is to expand it using $p\to q \equiv \neg p \lor q$. Then you get
$$(\forall x \in \mathbb{Z})(\exists y \in \mathbb{Z}) (2x + y \neq 3 \lor x + 2y = 3)$$
and it should be easiler to convince yourself that there is always an $y$ that makes $2x+y\neq 3$ hold ...
A: $A\to B$ is equivalent to $((\neg A)\lor B).$ So if $A$ is false then $A\to B$ is true. So $(\neg A) \implies (\;(\neg A)\land((\neg A)\lor B)\;)\implies (A\to B).$
So if $A$ is $2x+y=3,$  and $B$ is anything, then $\forall x\in \Bbb Z\;\exists y\in \Bbb Z\;(A\to B)$ is true if $\forall x\in \Bbb Z\;\exists y\in \Bbb Z\;(\neg A)$ is true.
In detail $$\forall x\in \Bbb Z\; \exists y\in \Bbb Z\;(y\ne 3-2x)\implies$$ $$\implies \forall x\;\exists y\in \Bbb Z\;(\neg A)\implies$$ $$\implies \forall x\in \Bbb Z\;\exists y\in \Bbb Z\;(\neg  A\land (A\to B))\implies$$ $$\implies \forall x\in \Bbb Z\;\exists Y\in \Bbb Z\;(A\to B).$$
