99 lines
2.1 KiB
Coq
99 lines
2.1 KiB
Coq
(* 22/09/2017. Someone asked during the course whether [~ (even 1)] can be
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proved, and if so, how. Here are several solutions, courtesy of
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Pierre-Evariste Dagand. *)
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Inductive even: nat -> Prop :=
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| even_O:
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even 0
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| even_SS:
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forall n, even n -> even (S (S n)).
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(* 1. The shortest proof uses the tactic [inversion] to deconstruct the
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hypothesis [even 1], that is, to perform case analysis. The tactic
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automatically finds that this case is impossible, so the proof is
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finished. *)
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Lemma even1_v1:
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even 1 -> False.
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Proof.
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inversion 1.
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(* In case you wish the see the proof term: *)
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(* Show Proof. *)
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Qed.
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(* For most practical purposes, the above proof *script* is good enough, and
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is most concise. However, those who wish to understand what they are doing
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may prefer to write a proof *term* by hand, in the Calculus of Inductive
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Constructions, instead of letting [inversion] construct a (possibly
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needlessly complicated) proof term. *)
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(* 2. Generalizing with equality. *)
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Lemma even1_v2':
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forall n, even n -> n = 1 -> False.
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Proof.
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exact (fun n t =>
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match t with
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| even_O =>
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fun (q: 0 = 1) =>
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match q with (* IMPOSSIBLE *) end
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| even_SS n _ =>
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fun (q : S (S n) = 1) =>
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match q with (* IMPOSSIBLE *) end
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end
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).
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Qed.
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Lemma even1_v2:
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even 1 -> False.
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Proof.
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eauto using even1_v2'.
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Qed.
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(* 3. Type-theoretically, through a large elimination. *)
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Lemma even1_v3':
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forall n,
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even n ->
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match n with
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| 0 => True
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| 1 => False
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| S (S _) => True
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end.
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Proof.
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exact (fun n t =>
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match t with
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| even_O => I
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| even_SS _ _ => I
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end
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).
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Qed.
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Lemma even1_v3:
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even 1 -> False.
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Proof.
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apply even1_v3'.
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Qed.
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(* 3'. Same technique, using a clever [match ... in ... return]. *)
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Lemma even1_v4':
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even 1 -> False.
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Proof.
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exact (fun t =>
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match t in even n
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return (
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match n with
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| 0 => True
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| 1 => False
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| S (S _) => True
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end
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(* BUG: we need the following (pointless) type annotation *)
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: Prop)
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with
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| even_O => I
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| even_SS _ _ => I
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end
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).
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Qed.
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