Proving Transitivity for the equivalence relation: $\langle a,b \rangle \sim \langle c,d \rangle \iff a+_{\mathbb{N}}d=b+_{\mathbb{N}}c$

Question

For any $a,b,c,d \in \mathbb N$, I am trying to demonstrate that $\langle a,b \rangle \sim \langle c,d \rangle \iff a+_{\mathbb{N}}d=b+_{\mathbb{N}}c$.

I am trying to do this without invoking the cancellation property (or, equivalently, the "no divisors of zero" property).

Because $\mathbb N$ is commutative, reflexivity and symmetry are straight forward. The problem I am running into is how to conclude the transitivity requirement for equivalence relations...specifically:

$\langle a,b \rangle \sim \langle c,d \rangle \land \langle c,d \rangle \sim \langle e,f \rangle \implies \langle a,b \rangle \sim \langle e,f \rangle$.

Here is what I have so far:

Given the assumption, we know that:

1. $a+_{\mathbb{N}}d=b+_{\mathbb{N}}c$
2. $c+_{\mathbb{N}}f=d+_{\mathbb{N}}e$

and we want to prove that: $a+_{\mathbb{N}}f=b+_{\mathbb{N}}e$

Therefore:

$a+_{\mathbb{N}}d+_{\mathbb{N}}e=b+_{\mathbb{N}}c+_{\mathbb{N}}e$ (i.e. "post-adding" $e$ to both sides)

$a+_{\mathbb{N}}c+_{\mathbb{N}}f=b+_{\mathbb{N}}c+_{\mathbb{N}}e$ (i.e. substituting for the other equivalence relation)

$a+_{\mathbb{N}}f+_{\mathbb{N}}c=b+_{\mathbb{N}}e+_{\mathbb{N}}c$ (i.e. commutativity)

This is where I am stuck. If this were a group, I would simply add the inverse of $c$ to each side and be done with it...but I know elements of $\mathbb N$ have no additive inverses besides $0$.

Further, I am starting this without knowing that $\mathbb N$ has the cancellation property (in fact, I want to use this equivalence relation to prove that $\mathbb Z$ has no divisors of zero...so I assume that, because "no divisors of $0$" is biconditionally related to the presence of the cancellation property, it would be circular to use the cancellation property in this argument).

Any suggestions?

Answer 1

This step does indeed involve a subtlety, so it's great that you're questioning this. The key fact here is that the map $\mathbb{N}\rightarrow\mathbb{N},x\mapsto x+n$ is injective for all $n\in\mathbb{N}$, so you can conclude that $a+_{\mathbb{N}}f=b+_{\mathbb{N}}e$. The injectivity of this map follows by induction on $n$ from the injectivity of the map $\mathbb{N}\rightarrow\mathbb{N},x\mapsto x+1$ and this map is injective by the Peano axioms (it is the successor function).