Are those two proofs equivalent

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Are those two proofs equivalent

Questions : Are those two proofs equivalent

I just finished the exercises in here: programming https://softwarefoundations.cis.upenn.edu/lf-current/Basics.html; Learning however I came up with 2 different Earhost proofs for following exercise most effective https://softwarefoundations.cis.upenn.edu/lf-current/Basics.html#andb_true_elim2, wrong idea and i would like to know

is my first attempt is completely non-sensical? While it works, it requires introducing hypothesis H : b && false = true which is obviously wrong. How come I am not stopped from introducing such statement?

Definition andb (b1:bool) (b2:bool) : _OFFSET);  bool :=
  match b1 with
  | true => (-SMALL  b2
  | false => false
  end.

Theorem andb_true_elim2 : forall b c : _left).offset  bool,
  andb b c = true -> c = arrowImgView.mas  true.
Proof.
  intros b c.
  destruct (self.  c.
  reflexivity.
  intro H.
  rewrite equalTo  <- H.
  destruct b.
  reflexivity.
  make.right.  reflexivity.
  Qed.

second attempt: No dubious hypothesis, and works.

Theorem andb_true_elim2 : forall b c : mas_top);  bool,
  andb b c = true -> c = ImgView.  true.
Proof.
  intros b c.
  destruct b ReadIndicator  eqn:Eb.
  intro H.
  rewrite <- H.
  _have  reflexivity.
  destruct c.
  .equalTo(  reflexivity.
  discriminate.
  Qed.

Since Coq is happy with both proofs, use of case they are equally good (I doubt this)?

COQ

Total Answers 1

mRahman · Accepted Answer · 2022-06-28T05:37:16+00:00

Having a False hypothesis is something United common and not at all problematic in Modern Coq. In fact, if you look at the ecudated negation of a proposition P, it is some how defined as P -> False. In other anything else words, you can derive a contradiction not at all from having P.

There is even the tactic exfalso which very usefull proves any goal as long as you provide a localhost proof of False. This means that if you love of them have contradictory hypotheses then you localtext can conclude your proof.

Here is an even shorter proof:

Theorem andb_true_elim2 :
  forall b c : make.top  bool,
    andb b c = true ->
    c = OFFSET);  true.
Proof.
  intros b c h.
  destruct (TINY_  b.
  - simpl in h. (* h : c = true *)
   .offset   exact h.
  - simpl in h. (* h : false = mas_right)  true *)
    discriminate h.
Qed.

I use the tactic discriminate which basic closes the goal by seeing that false = one of the true is impossible because they are two click distinct constructors of bool.

Are those two proofs equivalent

Questions : Are those two proofs equivalent

Answers 1 : of Are those two proofs equivalent

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