The axioms are given by:
$(∀x)¬(x^+ = 0)$
$(∀x)(∀y)(x+ = y^+ ⇒ x = y)$
Schema: for every expression P and variable symbol x, $P( 0) ⇒ (∀x)(P(x) ⇒ P(x^+)) ⇒ (∀x)P(x)$
$(∀x)(x + 0 = x)$
$(∀x)(∀y)(x + y^+ = (x + y)^+) $
$(∀x)(x.0 = 0)$
$(∀x)(∀y)(x.(y^+) = (x.y) + x)$
And $<$ is defined as an abbreviation for $(∃u)(x + u^+ = y)$
I am asking to prove that $(∀m)(∀ n)(0 < n ⇒ (∃!q)(∃r)(m = q·n + r∧r < n)) $ using the above axioms, but can hardly think of where to get start. Could someone please give a proof? Thanks in advance!
Wow, I really hope that this is not your first proof you have to give on the basis of the Peano Axioms, because this is not easy ... typically a first proof would be something much simpler, like $\forall x \forall y x + y = y + x$. So if you're a beginner with these Peano Axioms, I can understand that you're at a bit of a loss as to how to even start this.
OK, before trying a formal proof on the basis of the Peano Axioms, it makes sense to first get straight on a mathematical proof of why this is so. Here is such a proof:
First, let's prove that there definitely does exist at least one $q$ (and $r$) with the desired property. We can do this by induction over $m$:
Base: $m=0$. For any $n$, the following holds: $0 = 0 \cdot n + 0$, and of course $0 < n$. That is, we can pick $q = 0$ and $r = 0$.
Step: Let $m$ be arbitrary, and suppose (inductive hypothesis) that for any $n$ there exists a $q$ for which there is an $r$ such that $m = q\cdot n + r \land r < n$. Now let's show that for any $n$ there is a $q'$ for which there is an $r'$ such that $m^+ = q'\cdot n + r' \land r' < n$.
There are really only two cases to consider
Well, if $r^+ < n$, then $m^+ = q \cdot n + r^+ \land r^+ < n$, so set $q' = q$ and $r' = r^+$
If $r^+ = n$, then $m^+ = q^+ \cdot n + 0 \land 0 < n$, so set $q' = q^+$ and $r' = 0$
OK, so this proves existence. Now let's prove uniqueness. We do this using a proof by contradiction:
Suppose you have an $m$ and $n$, $q_1$, and $r_1$, $q_2$, and $r_2$ such that:
$m = q_1 \cdot n + r_1 \land r_1 < n$
$m = q_2 \cdot n + r_2 \land r_2 < n$
and where $q_1 \not = q_2$
Since $q_1 \not = q_2$, we have $q_1 < q_2$ or $q_1 > q_2$
Let's just consider the case $q_1 < q_2$ (the other one is analogous).
So that means $q_2 = q_1 + d$ for some $d > 0$
Since we have that $m = q_2 \cdot n + r_2$
We thus get: $m = (q_1 + d) \cdot n + r_2$
And therefore:
$m = q_1 \cdot n + d \cdot n + r_2$
But since we had that $m = q_1 \cdot n + r_1$, it must be that:
$r_1 = d \cdot n + r_2$
But since $d>0$, we thus get $r_1 \ge n$, which contradicts $r_1 < n$
So, we can't have multiple $q$'s (and $r$'s) for which the equation holds, so that proves uniqueness.
QED
OK, so now that you know what proof to formalize, you still have a ways to go, but at least you have a proof plan. Now, I would try to first formalize and prove some of the supporting Lemma's. Here are just two of them:
$\forall x \forall y (x < y \lor x = y \lor y < x)$ (Trichotomy)
$\forall x \forall y \forall z (x + y = x + z \rightarrow y = z)$ (Addition (Left) Cancellation)
Both of these are doable to prove from the PA axioms, but still not easy (you'll need to use the induction schema for both of them). But once you have them, the above proof should follow fairly easily, though you want to be careful how to lay out the proof by contradiction for the uniqueness of $q$ inside the universal proof for $m$ and $n$.
Good luck!!