Help with proposition whether it's true or false Is this proposition true or false?
$$\exists y \in \mathbb R \;\forall x \in \mathbb R\,(xy\neq x \rightarrow x=0) $$
And why?  
I'm confused as to what exactly is being claimed. 
 A: It is true. $y=1\in\mathbb R$ does the job. 
Note that $x\neq x\Rightarrow x=0$ is actually the same statement as $x=x\vee x=0$. 
This statement is evidently true for each $x\in\mathbb R$.
A: Take $y = 1$ and show that $x \neq x \implies x = 0\tag{1}$
Note that you can express this in its logically equivalent contrapositive form, which might help you see that the statement is in fact true.
Take $y = 1$. Then we have $x\neq 0 \implies x = x$, which is certainly true, because $x = x$ is always true. So no matter whether $x = 0$ or $x\neq 0$, the implication must be true.
Recall: an implication $p \rightarrow q$ is true whenever $p$ is false or $q$ is true.
A: One way to examine whether a "there exists" statement is trus is to see whether it's true for some particular values. Suppose you say "If I pick $y = 1$, is the rest true?" You might have a little trouble answering that, so move on. 
Try saying "If I pick $y = 2$, is the remainder true?"  What result do you get? 
A: The earlier answers all pull $\;y=1\;$ as a rabbit out of a hat.  Let me show you a way to avoid that.

$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\then}{\mathop{\;\Rightarrow\;}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$Implicitly letting $\;x\;$ and $\;y\;$ range over $\;\mathbb R\;$ to reduce clutter, and using a slightly different notation, we calculate
$$\calc
\langle \forall x :: xy \not= x \then x = 0 \rangle
\calcop\equiv{logic: write $\;P \then Q\;$ as $\;\lnot P \lor Q\;$ -- to get rid of the $\;\not=\;$}
\langle \forall x :: xy = x \;\lor\; x = 0 \rangle
\calcop\equiv{arithmetic: divide both sides of $\;xy = x\;$ by $\;x\;$, special case $\;x = 0\;$}
\langle \forall x :: x = 0 \lor y = 1 \;\lor\; x = 0 \rangle
\calcop\equiv{logic: simplify; pull part not using $\;x\;$ out of $\;\forall x\;$}
y = 1 \;\lor\; \langle \forall x :: x = 0 \rangle
\calcop\equiv{arithmetic/logic: not all real numbers are 0; simplify}
y = 1
\endcalc$$
From this, and the rule of logic that says that $\;\langle \exists y :: y = \dots\rangle\;$ is true, it immediately follows that your proposition is true.

Observe also how the above calculation doesn't just show that $\;y = 1\;$ 'does the job': the equivalences show that it is the only such real number.
