自然数中非等于转相等及Coq函数非零相关定理改写问询

Alright, let’s work through this problem step by step. You’ve already confirmed that the result of f1 + m is a non-zero natural number (meaning it’s guaranteed to be of the form S k for some k), and now you need to formalize the leap from "not equal to 0" to "exactly equal to S m" in your target theorem. Here’s a concrete, actionable approach:

1. First, Map Boolean Inequalities to Propositional Ones #

Your theorem uses =? (the boolean equality check) with false as the result. To work with Coq’s propositional logic, we’ll convert these boolean statements to actual inequality propositions using the built-in Nat.eqb_neq lemma, which states that x=?y = false is equivalent to x ≠ y.

2. Core Proof Structure #

Here’s the full proof scaffolding, with comments explaining each step. Note that we’ll need to rely on properties of your f1 function—since without constraints on f1, f1 + m = S m isn’t universally true (e.g., if f1=2 and m=1, 2+1=3≠S1=2). I’ll assume you have a pre-proven lemma about f1 (like f1 = 1 or a similar constraint) to fill in the final gap.

Require Import Nat.

-- Assume `f1` is your pre-defined function (adjust this to match your actual definition)
Variable f1 : nat.

Theorem greater : forall (m : nat) (l : list nat),
  m=?0 = false ->
  0=?(f1 + m) = false ->
  (f1 + m) = S m.
Proof.
  -- Introduce all variables and premises into the context
  intros m l Hm_nonzero Hsum_nonzero.

  -- Convert boolean "not equal" to propositional inequalities
  apply Nat.eqb_neq in Hm_nonzero.  (* Now Hm_nonzero : m ≠ 0 *)
  apply Nat.eqb_neq in Hsum_nonzero. (* Now Hsum_nonzero : f1 + m ≠ 0 *)

  -- Since `f1 + m ≠ 0`, it must be of the form `S k` (by natural number induction)
  destruct (f1 + m) as [|k].
  - (* Case 1: f1 + m = 0 *)
    contradiction Hsum_nonzero.  -- This contradicts our `f1 + m ≠ 0` premise, so discard.
  - (* Case 2: f1 + m = S k *)
    -- Now we need to show k = m, which means f1 + m = S m.
    -- Here’s where you plug in your pre-proven lemma about `f1`.
    -- Example: If you know `f1 = 1`, use these steps:
    -- rewrite <- Nat.add_1_r.  -- Rewrites S m to m + 1
    -- rewrite Nat.add_comm.    -- Swaps f1 and m if needed
    -- apply f1_eq_1.           -- Where `f1_eq_1 : f1 = 1` is your pre-proven lemma
    -- reflexivity.             -- Closes the goal once everything matches
Admitted.  -- Replace `Admitted` with your actual proof steps for `f1`

3. Key Dependencies to Finish the Proof #

The missing piece here is a lemma defining f1’s behavior. For example:

• If f1 always returns 1, prove f1_eq_1 : f1 = 1 and use it to rewrite f1 + m to 1 + m, which equals S m by Nat.add_1_r.
• If f1 is derived from the list l (e.g., f1 is the length of l), you’ll need to use properties of lists (like length l = 1 in this case) to constrain f1.

4. Why This Works #

• We use destruct to split the natural number f1 + m into its only two possible forms: 0 or S k. The 0 case is impossible thanks to your Hsum_nonzero premise.
• Once we know f1 + m = S k, we just need to link k to m—which relies on f1’s fixed value or behavior that you’ve already reasoned about.

内容的提问来源于stack exchange，提问作者rosi javi