Rev.v le_antisymmetric

Question

I came to this point:

Theorem le_antisymmetric :
  antisymmetric le.
Proof.
  unfold antisymmetric. intros a b H1 H2. generalize dependent a.
  induction b as [|b' IH].
  - intros. inversion H1. reflexivity.
  - intros.

Output:

b' : nat
IH : forall a : nat, a <= b' -> b' <= a -> a = b'
a : nat
H1 : a <= S b'
H2 : S b' <= a
------------------------------------------------------
a = S b'

My plan was to use transitivity of le:

a <= b -> b <= c -> a <= c

And substitute a := a, b := (S b') and c := a.

So we'll get:

a <= (S b') -> (S b') <= a -> a <= a

I'll use H1 and H2 as 2 hypotheses needed and get Ha: a <= a. Then do an inversion upon it, and get the only way construct this is a = a.

But what syntax should I use to apply transitivity with 2 my hypotheses to get Ha?

`Goal forall x y, x <= y -> y <= x -> x = y. induction x; destruct y; subst; [reflexivity | inversion 2 | inversion 1 | intros; apply f_equal, IHx; apply le_S_n; trivial]. Qed.` — larsr, Sep 17 '19 at 06:53

score 2 · Accepted Answer · answered Sep 15 '19 at 22:05

Your first induction over b here seems unnecessary. Consider le:

Inductive le (n : nat) : nat -> Prop :=
    le_n : n <= n | le_S : forall m : nat, n <= m -> n <= S m

You should instead be inspecting H1 first. If it's le_n, then that's equality, and you're done. If it's le_S, then presumably that's somehow impossible.

intros a b [ | b' H1] H2.
- reflexivity.

This leaves us with

a, b, b' : nat (* b is extraneous *)
H1 : a <= b'
H2 : S b' <= a
______________________________________(1/1)
a = S b'

Now, transitivity makes sense. It can give you S b' <= b', which is impossible. You can derive a contradiction using induction (I think), or you can use an existing lemma. The whole proof is thus.

intros a b [ | b' H1] H2.
- reflexivity.
- absurd (S b' <= b').
  + apply Nat.nle_succ_diag_l.
  + etransitivity; eassumption.

That last bit is one way to use transitivity. etransitivity turns the goal R x z into R x ?y and R ?y z, for a new existential variable ?y. eassumption then finds assumptions that match that pattern. Here, specifically, you get goals S b' <= ?y and ?y <= b, filled by H2 and H1 respectively. You can also give the intermediate value explicitly, which lets you drop the existential prefix.

transitivity a; assumption.

I used `apply le_Sn_n.` instead of `apply Nat.nle_succ_diag_l.` — user4035, Sep 16 '19 at 07:57

Rev.v le_antisymmetric

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