在Coq中,假设我有一个f在(g x)上匹配定义的fixpoint函数,我想g x = ...在证明中使用form()中的假设.以下是一个最小的工作示例(实际上f,g会更复杂):
Definition g (x:nat) := x.
Fixpoint f (x:nat) :=
match g x with
| O => O
| S y => match x with
| O => S O
| S z => f z
end
end.
Lemma test : forall (x : nat), g x = O -> f x = O.
Proof.
intros.
unfold f.
rewrite H. (*fails*)
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该消息显示Coq被卡住的地方:
(fix f (x0 : nat) : nat :=
match g x0 with
| 0 => 0
| S _ => match x0 with
| 0 => 1
| S z0 => f z0
end
end) x = 0
Error: Found no subterm matching "g x" in the current goal.
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但是,这些命令unfold f. rewrite H.不起作用.
我如何获得Coq unfold f然后使用H?
Parameter g: nat -> nat.
(* You could restructure f in one of two ways: *)
(* 1. Use a helper then prove an unrolling lemma: *)
Definition fhelp fhat (x:nat) :=
match g x with
| O => O
| S y => match x with
| O => S O
| S z => fhat z
end
end.
Fixpoint f (x:nat) := fhelp f x.
Lemma funroll : forall x, f x = fhelp f x.
destruct x; simpl; reflexivity.
Qed.
Lemma test : forall (x : nat), g x = O -> f x = O.
Proof.
intros.
rewrite funroll.
unfold fhelp.
rewrite H.
reflexivity.
Qed.
(* 2. Use Coq's "Function": *)
Function f2 (x:nat) :=
match g x with
| O => O
| S y => match x with
| O => S O
| S z => f2 z
end
end.
Check f2_equation.
Lemma test2 : forall (x : nat), g x = O -> f2 x = O.
Proof.
intros.
rewrite f2_equation.
rewrite H.
reflexivity.
Qed.
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