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another problem from SFv1 which got me stuck.

The theorem is as follows:

Theorem plus_le_compat_l : forall n m p,
  n <= m ->
  p + n <= p + m.

I have tried several avenues so far, the one that got me the furthest was to introduce n and start induction on it. The base case is rather trivial (intros m p H. rewrite add_0_r. apply le_plus_l.).

As for the inductive step, I have tried to destruct p, which gives me two subcases. For the first where p = O, it's super easy to prove by applying H. When p = S n I get stuck. Here's the full proof so far:

Theorem plus_le_compat_l : forall n m p,
  n <= m ->
  p + n <= p + m.
Proof.
  intros n.
  induction n as [| n' IHn'].
  - intros m p H. rewrite add_0_r. apply le_plus_l.
  - destruct p eqn:E.
    + intros H. simpl. apply H.
    + intros H. simpl. apply n_le_m__Sn_le_Sm.

And my current goal:

  n' : nat
  IHn' : forall m p : nat, n' <= m -> p + n' <= p + m
  m, p, n : nat
  E : p = S n
  H : S n' <= m
  ============================
  n + S n' <= n + m

I tried a trivial manipulation to turn the LHS of the inequality into n' + S n which lets me rewrite the S n as p, but that also doesn't take me anywhere.

Any hints here are highly appreciated :-)

Thanks

PS: I have tried to apply a previously proved theorem without success:

Theorem add_le_cases : forall n m p q,
  n + m <= p + q -> n <= p \/ m <= q.
Felipe Balbi
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1 Answers1

1

I think in this case using induction on p is the way to go, and both goals should be quite ok to prove – the induction one by a use of le_n_S.

Meven Lennon-Bertrand
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