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程序修复点的Coq简化

有喜欢的战术什么simplProgram FixpointS'

特别是,如何证明以下琐碎的陈述?

Program Fixpoint bla (n:nat) {measure n} :=
match n with
| 0 => 0
| S n' => S (bla n')
end.

Lemma obvious: forall n, bla n = n. 
induction n. reflexivity.
(* I'm stuck here. For a normal fixpoint, I could for instance use 
simpl. rewrite IHn. reflexivity. But here, I couldn't find a tactic 
transforming bla (S n) to S (bla n).*)
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显然,Program Fixpoint这个玩具示例没有必要,但我在一个更复杂的设置中面临同样的问题,我需要证明Program Fixpoint手动终止.

theorem-proving coq totality

9
推荐指数
1
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698
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coq ×1

theorem-proving ×1

totality ×1