Coq字符串语法定理证明受阻：归纳序选择困惑求助

First, let's diagnose why your initial induction attempts struggled: when you induct directly on a, b, or stream, you end up with overly general induction hypotheses that don't tie the structure of the token path to the token stream via the PathMatchesStream relation. The key insight here is to induct on the PathMatchesStream proof term itself—this lets you leverage the recursive structure of the matching relation, which directly links the shape of the path and stream.

Core Proof Idea #

The theorem states that if two token paths match the same stream, they must be identical. Since PathMatchesStream is defined recursively by building paths and streams from their heads, we can mirror that recursion in our proof:

1. For base cases where the path/stream are single elements, we use your existing TokenDefinition_match_same_then_same theorem to show the single definitions are equal.
2. For recursive cases where paths/streams are built by prepending elements, we first show the head definitions are equal, then apply the induction hypothesis to the tail paths/streams.

Step-by-Step Proof Code #

Here's a working implementation using your existing tactics and definitions:

Theorem PathMatchesStream_same_if_match_same: forall a b stream, PathMatchesStream a stream -> PathMatchesStream b stream -> a = b.
Proof.
  intros a b stream Ha Hb.
  (* Induct on the structure of the PathMatchesStream proof for a and stream *)
  induction Ha; inversion Hb; clear Hb.
  - (* Case 1: Ha is PathMatchesStream_base, Hb is also PathMatchesStream_base *)
    simpl_TokenMatches.
    crush. (* Uses TokenDefinition_match_same_then_same under the hood *)
  - (* Case 2: Ha is append, Hb is base → contradiction (stream length mismatch) *)
    invert_PathMatchesStream.
  - (* Case 3: Ha is base, Hb is append → contradiction (stream length mismatch) *)
    invert_PathMatchesStream.
  - (* Case 4: Both Ha and Hb are append constructions *)
    simpl_TokenMatches.
    f_equal. (* Reduce list equality to head equality + tail equality *)
    + (* Prove head definitions are equal *)
      apply TokenDefinition_match_same_then_same; assumption.
    + (* Apply induction hypothesis to the tail paths/streams *)
      apply IHa; assumption.
Qed.

Key Explanations #

• Inducting on Ha: By inducting on the proof of PathMatchesStream a stream, we get an induction hypothesis that directly applies to the sub-path a' and sub-stream stream'—this is far more useful than inducting on the list a alone, as it preserves the critical link between the path and stream.
• inversion Hb: This tactic automatically splits Hb into cases that align with Ha's structure. Any mismatched cases (like Ha being an append and Hb being a base) are immediately contradictory, and your invert_PathMatchesStream tactic handles those contradictions using your pre-proven length theorems.
• f_equal: To prove two lists are equal, we just need to show their heads are equal and their tails are equal. The head equality comes from your existing token matching theorem, and the tail equality comes directly from our induction hypothesis.

Why Your Previous Attempts Failed #

When you inducted on a, b, or stream separately, you ended up with induction hypotheses that didn't account for the PathMatchesStream relation. For example, inducting on a gives you a hypothesis like forall b stream, PathMatchesStream a' stream -> PathMatchesStream b stream -> a' = b—this is too broad, because it doesn't tie a' to the specific stream' that's the tail of the original stream. Inducting on the PathMatchesStream proof fixes this by linking the sub-path and sub-stream in the hypothesis.

内容的提问来源于stack exchange，提问作者blaineh