Are there "interesting" theorems in Peano arithmetic, that only use the addition operation?

Question

More precisely, are there "interesting" theorems of Presburger arithmetic, other than the following four well-known "interesting" ones?

1. The commutativity of addition.

2. The theorem stating there are no two consecutive even numbers.

3. The theorem stating that every natural number, either is even or is followed by an even successor (along with analogous theorems about a sum of more than two identical natural numbers, e.g., the theorem stating that every natural number not being any sum of three identical natural numbers - is followed by a successor that either is such a sum or is followed by a successor which is such a sum).

4. $\forall(a,c)\exists(b)[(a+b=c)\lor (c+b=a)]$

For our purposes, for an "additive" theorem to be "interesting" enough: 1) it must have an intuitive meaning [like that of the theorems I've mentioned as examples]; 2) its proof should need to rely on induction; 3) It should not be an intuitive generalization of any axiom [e.g. the axiom x+(y+1)=(x+y)+1, which can be viewed as a special case wherein z=1, intuitively generalized - for all z - as the theorem of associativity of addition: x+(y+z)=(x+y)+z].

Answer 1

For fixed relatively prime integers $m$ and $n$, the fact that a number divisible by both of them is divisible by their product is expressible in Presburger arithmetic. The case $m=2$, $n=3$:

$$\forall x[(\exists y)(y+y=x)\land(\exists y)(y+y+y=x)\to(\exists y)(y+y+y+y+y+y=x)]$$

Answer 2

The associativity of addition is one candidate. Recall from the definition of addition that \begin{align} a+0 &=a & \text{A1} \\ a+S(b) &=S(a+b) & \text{A2} \end{align}

Theorem: $(a+b)+c = a+(b+c)$

Proof: Let $\phi(c)$ denote the proposition that $(a+b)+c = a+(b+c)$. Proceed by induction on $c$.

Base case: Let $c=0$. Then \begin{align} (a+b)+c &= (a+b)+0 && \text{A1} \\ &= a+b \\ &= a+(b+0) && \text{A1} \\ &= a+(b+c) \end{align}

Therefore $\phi(0)$.

Inductive step: Suppose $(a+b)+c = a+(b+c)$. Then \begin{align} (a+b)+S(c) &= S((a+b)+c) && \text{A2} \\ &= S(a+(b+c)) && \text{inductive hypothesis} \\ &= a+S(b+c) && \text{A2} \\ &= a+(b+S(c)) && \text{A2} \end{align}

Therefore $\phi(c) \rightarrow \phi(c+1)$ for all $c \in \mathbb{N}$.

By the axiom schema of induction, $\phi(c)$ for all $c \in \mathbb{N}$. This proves that $(\mathbb{N}, +)$ is a semigroup. Together with the proof that 0 is a left identity (in addition to being a right identity as stated in A1), this proves that $(\mathbb{N}, +)$ is a monoid.

Answer 3

Presburger Arithmetic can express $x\equiv y\pmod n$ for fixed $n$, so we can state instances of the Chinese Remainder Theorem for fixed moduli, such as

For all $a$ and $b$ there is an $x$ such that $x\equiv a\pmod{7}$ and $x\equiv b\pmod{8}$.

Since this claim can be expressed and it's true in $\mathbb N$, Presburger Arithmetic, being complete, must be able to prove it.