Concluding that $x=0$ from $m+x=m$ I am studying a book on proofs and there are two statements that I don't understand the difference:


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*Let $x$ belong to the set of integers. If $x$ has the property that for each integer $m$, $m + x = m$, then $x = 0$.

*Let $x$ belong to the set of integers. If $x$ has the property that there exists an integer $m$ such that $m + x = m$, then $x = 0$. 
I would greatly appreciate the community's input, so that I can focus on the proofs. For statement 1, how about, since m belongs to $\mathbb{Z}$:
$$ m + x = m  $$
 $$(m + x) = (m)  $$
 $$ (-m) + (m + x) = (-m) + (m) $$ 
 $$ ((-m) + m) + x = 0 $$
$$ (0) + x = 0 $$ 
$$ x = 0$$
Would that be good? This would say that there is only one unique solution for any $m\in\mathbb{Z}$ and it's $0$. Moreover, for statement 2, it seems that the logic is the same, but I don't need to write it as variables. Instead, I simply need to pick one single value, right?  
 A: 2 has a weaker condition. In 2, the only thing we need to show that $x = 0$ is that there is a single integer $m$ such that $x + m = m$. For 1, however, it must hold for all integers, $m$, that $x + m = m$.
A: In the first statement, $m+x=m$ for all integers $m$.  In the second statement, we only need $x+m=m$ for at least one integer $m$.
A: Statement 1 requires $m + x = m$ for all integer $m$, so that $m$ could be $-197$, or $3^{45} + 2$, or any integer whatsoever out of infinitely many. You have to prove that this works for any integer value of $m$ that could possibly be chosen.
Statement 2 only needs a single instance of $m$ such that $m + x = m$. You only need to produce one value of $m$ that gives the desired result. You could assign $m = 47$ (to pick a number out of a hat) and then show that the only solution to $47 + x = 47$ is $x = 0$.
As it happens, both statements are true, and 0 is sometimes called "the additive identity" (and in fact this $m + x = m$ thing also works for $m$ being any real, imaginary or complex number). And although both statements are easy to accept without proof (some might consider them to be axioms), statement 2 is much easier to prove than statement 1.
