"For all" versus "There exists"

Question

My class uses the book "The Art of Proof" by Matthias Beck and Ross Geoghegan. Proposition 1.12 states: Let $x\in\mathbb{Z}$. If $x$ has the property that for each integer $m$, $m+x=m$, then $x=0$.

The proof isn't shown in the book, but the proof in class went like this:

Let $x\in\mathbb{Z}$. Let $x$ have the property that for each $m\in\mathbb{Z}$, $m+x=m$.

If we choose $m=0$, then $x=0=x+0$, as $0$ is the additive identity.

There is a similar yet different Proposition 1.13 that states:

Let $x\in\mathbb{Z}$. If $x$ has the property that there exists an integer $m$ such that $m+x=m$, then $x=0$.

The proof went like this:

Let $x\in\mathbb{Z}$. Let $x$ have the property that there is $m\in\mathbb{Z}$ such that $m+x=m$. Further let $m$ be this number.

$m+x=m+0$ by Axiom 1.2 (Additive identity)

$x=0$ by Proposition 1.9 (not important to the question)

I'm very confused as to why Proposition 1.12, using "for each", uses a specific $m$ while Proposition 1.13, which uses "there exists", doesn't make assumptions about $m$. Shouldn't it be the other way around? Doesn't "for each" imply a burden of proof for every integer while "there exists" means that there is at least one integer that fits the equation? I can't find any information about "for each" versus "there exists" in the textbook or online.

Answer 1

Doesn't "for each" imply a burden of proof for every integer while "there exists" means that there is at least one integer that fits the equation?

Yes, but let us look at the statement of Proposition 1.12 carefully:

Let $x \in \mathbb{Z}$. If $x$ has the property that for each integer $m$, $m+x=m$ then $x=0$.

What are the assumptions and what are we trying to prove? There are two assumptions here:

Assumption 1: $x \in \mathbb{Z}$.

Assumption 2: $x$ has the property that for each integer $m$, $m+x=m$.

What are we trying to prove?

To prove: $x = 0$.

Notice that the "for each" statement appears in an assumption, not in the statement we are trying to prove! So even though it is true that "'for each' impl[ies] a burden of proof for every integer", that is irrelevant for the current proof, because we are not trying to prove a 'for each' statement. We are assuming a 'for each' statement, i.e., we already know that assumption 2 holds for each integer $m$. Therefore, we are allowed to substitute any integer $m$ into assumption 2.

On the contrary, if have an assumption starting with "there exists $x$ such that blah blah", then we cannot substitute an arbitrary value of $x$ into "blah blah", because our assumption only guarantees that there is some value $x$ such that "blah blah" holds.

So it is exactly the opposite depending on whether we are talking about assumptions, or the statement to be proved.

If we are trying to prove a "for each" statement, then we have a burden of proof for every value.

If we are trying to prove a "there exists" statement, then we have a burden of proof for only one value.

If we have an assumption that is a "for each" statement, then we may substitute any value into the variable.

If we have an assumption that is a "there exists" statement, then we may not make an arbitrary substitution; we can only use the variable as it is, since we do not know anything else about it.