# Injectivity of succ

In the last post, I showed how to prove, using only elementary tools, that zero is not the successor of any natural number, that `O = succ(m)` leads to a contradiction.

In this post, we’ll see that the successor function is injective. Writing `S` in place of `succ`, we’ll show that if `S(m) = S(n)`, then `m = n`.

As before, we start with our own version of the natural numbers.

``````inductive mynat : Type
| O : mynat
| S : mynat → mynat
``````

The strategy for our proof will be to find a predicate `P` on pairs of natural numbers such that on the assumption `h : S(m) = S(n)`, the following criteria are satisfied:

1. From `h₃`, a proof of `P (S m) (S n)`, we can easily prove the result `m = n`.
2. We can easily construct `h₁`, a proof of `S(m) = S(m) → P (S m) (S m)`.

Why does this help? Substituting equation `h` into `h₁` gives

`h₂ : S(m) = S(n) → P (S m) (S n)`. But we know `h`, a proof of the antecedent of this implication, and hence we derive `h₃ := h₂ h`, the desired proof of `P (S m) (S n)`. From this, we prove our initial target `m = n`.

I’m sure you will agree that the first condition above is satisfied if `P` is chosen so that `P (S m) (S n) = m = n`.

Since this is a good candidate for `P`, we’ll give it a Lean definition.

``````def P : mynat → mynat → Prop
| (S a) (S b) := a = b
| _     _     := true``````

The second pattern in the definition isn’t used on our proof, but Lean functions are total, so we must define `P` in all cases.

What remains is to prove the second criterion. By definition, `P (S m) (S m)` is simply `m = m`. This is true by reflexivity.

Those are all the ingredients. Time to assemble them up into a Lean proof.

``````lemma succ_inj (m n : mynat) (h : S m = S n) : m = n :=
have h₁ : S(m) = S(m) → P (S m) (S m),  from λ _, rfl,
have h₂ : S(m) = S(n) → P (S m) (S n),  from h ▸ h₁,
have h₃ : P (S m) (S n),                from h₂ h,
show m = n,                             from h₃``````