How do I prove that the addition of natural numbers, as defined by Peano, is unique?

Question

In other words, for any two natural numbers, there exist no more than one natural number that equals the sum of the two numbers. Or rather, for any two natural numbers, their sum is unique.

In first order logic, I am trying to prove the following:

$\forall a,b,c,c'\in \mathbb{N}(a+b=c \land a+b=c' \implies c=c')$

I tried induction on $b$, and proved the base case for $b=0$, but am having trouble with the inductive step.

Edit: I was confused while asking the question. I am indeed looking for a proof that addition is injective, rather than one that equality is transitive.

Answer 1

Suppose $a + b = c$ and $a + b = c’$. By the symmetric property of equality, $c = a + b$. By the transitive property of equality, $c = c’$.

As you can see, the proof here is trivial. So there is probably some source of confusion if you think the proof shouldn’t be trivial.

Answer 2

Equality is an equivalence relation. Equivalence relations are all defined by being reflexive, symmetric, and transitive. So by the transitive property of equality $c=a+b=c' \implies c=c'$. And this is also the $4^{\text{th}}$ Peano axiome.

Answer 3

I'd introduce addition as follows:

$a + 0 = a$ and $a + v(b) = v(a+b)$ where $v$ denotes the successor.

This leads to an answer of your problem.