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:
- From
h₃
, a proof ofP (S m) (S n)
, we can easily prove the resultm = n
. - We can easily construct
h₁
, a proof ofS(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
This is very nice, but is there a simpler way? I was assuming that injectivity of the constructors of an inductive type would be a foundational assumption of the system. Does it even make sense to have recursors defined on a type if it weren’t injective?
I looked into this more and now understand that constructors are not always injective: the exception is when the inductive type is a Prop type. Since we have proof-irrelevance, all values of a Prop type are equal, and so you can’t possibly extract the different elements from them. See Exists for a concrete use of that: you can’t pull out the particular element that exists from it.
So if mynat here were instead a Prop, what goes wrong? I couldn’t see it until I tried, but the problem is that rec_on for a Prop can only produce a Prop type. But the definition of P here is producing not a Prop type but Prop itself, which is of course a Type. That’s not allowed: we can only yield a Prop.
Thanks for the blog posts – they’re very enjoyable and helpful for beginners like myself at this.
Thanks Tom. I’m afraid I haven’t touched the blog in several months so only just saw your comment.