Don't understand `destruct` tactic on hypothesis `~ (exists x : X, ~ P x)` in Coq - coq

Question

I'm new to Coq and try to learn it through Software foundations. In the chapter "Logic in Coq", there is an exercise not_exists_dist which I completed (by guessing) but not understand:
Theorem not_exists_dist :
excluded_middle →
∀ (X:Type) (P : X → Prop),
¬ (∃ x, ¬ P x) → (∀ x, P x).
Proof.
intros em X P H x.
destruct (em (P x)) as [T | F].
- apply T.
- destruct H. (* <-- This step *)
exists x.
apply F.
Qed.
Before the destruct, the context and goal looks like:
em: excluded_middle
X: Type
P: X -> Prop
H: ~ (exists x : X, ~ P x)
x: X
F: ~ P x
--------------------------------------
(1/1)
P x
And after it
em: excluded_middle
X: Type
P: X -> Prop
x: X
F: ~ P x
--------------------------------------
(1/1)
exists x0 : X, ~ P x0
While I understand destruct on P /\ Q and P \/ Q in hypothesis, I don't understand how it works on P -> False like here.

Answer 1

Let me try to give some intuition behind this by doing another proof first.
Consider:
Goal forall A B C : Prop, A -> C -> (A \/ B -> B \/ C -> A /\ B) -> A /\ B.
Proof.
intros. (*eval up to here*)
Admitted.
What you will see in *goals* is:
1 subgoal (ID 77)
A, B, C : Prop
H : A
H0 : C
H1 : A ∨ B → B ∨ C → A ∧ B
============================
A ∧ B
Ok, so we need to show A /\ B. We can use split to break the and apart, thus we need to show A and B. A follows easily by assumption, B is something we do not have. So, our proof script now might look like:
Goal forall A B C : Prop, A -> C -> (A \/ B -> B \/ C -> A /\ B) -> A /\ B.
Proof.
intros. split; try assumption. (*eval up to here*)
Admitted.
With goals:
1 subgoal (ID 80)
A, B, C : Prop
H : A
H0 : C
H1 : A ∨ B → B ∨ C → A ∧ B
============================
B
The only way we can get to the B is by somehow using H1. Let's see what destruct H1 does to our goals:
3 subgoals (ID 86)
A, B, C : Prop
H : A
H0 : C
============================
A ∨ B
subgoal 2 (ID 87) is:
B ∨ C
subgoal 3 (ID 93) is:
B
We get additional subgoals! In order to destruct H1 we need to provide it proofs for A \/ B and B \/ C, we cannot destruct A /\ B otherwise!
For the sake of completeness: (without the split;try assumption shorthand)
Goal forall A B C : Prop, A -> C -> (A \/ B -> B \/ C -> A /\ B) -> A /\ B.
Proof.
intros. split.
- assumption.
- destruct H1.
+ left. assumption.
+ right. assumption.
+ assumption.
Qed.
Another way to view it is this: H1 is a function that takes A \/ B and B \/ C as input. destruct works on its output. In order to destruct the result of such a function, you need to give it an appropriate input.
Then, destruct performs a case analysis without introducing additional goals.
We can do that in the proof script as well before destructing:
Goal forall A B C : Prop, A -> C -> (A \/ B -> B \/ C -> A /\ B) -> A /\ B.
Proof.
intros. split.
- assumption.
- specialize (H1 (or_introl H) (or_intror H0)).
destruct H1.
assumption.
Qed.
From a proof term perspective, destruct of A /\ B is the same as match A /\ B with conj H1 H2 => (*construct a term that has your goal as its type*) end.
We can replace the destruct in our proof script with a corresponding refine that does exactly that:
Goal forall A B C : Prop, A -> C -> (A \/ B -> B \/ C -> A /\ B) -> A /\ B.
Proof.
intros. unfold not in H0. split.
- assumption.
- specialize (H1 (or_introl H) (or_intror H0)).
refine (match H1 with conj Ha Hb => _ end).
exact Hb.
Qed.
Back to your proof. Your goals before destruct
em: excluded_middle
X: Type
P: X -> Prop
H: ~ (exists x : X, ~ P x)
x: X
F: ~ P x
--------------------------------------
(1/1)
P x
After applying the unfold not in H tactic you see:
em: excluded_middle
X: Type
P: X -> Prop
H: (exists x : X, P x -> ⊥) -> ⊥
x: X
F: ~ P x
--------------------------------------
(1/1)
P x
Now recall the definition of ⊥: It's a proposition that cannot be constructed, i.e. it has no constructors.
If you somehow have ⊥ as an assumption and you destruct, you essentially look at the type of match ⊥ with end, which can be anything.
In fact, we can prove any goal with it:
Goal (forall (A : Prop), A) <-> False. (* <- note that this is different from *)
Proof. (* forall (A : Prop), A <-> False *)
split; intros.
- specialize (H False). assumption.
- refine (match H with end).
Qed.
Its proofterm is:
(λ (A B C : Prop) (H : A) (H0 : C) (H1 : A ∨ B → B ∨ C → A ∧ B),
conj H (let H2 : A ∧ B := H1 (or_introl H) (or_intror H0) in match H2 with
| conj _ Hb => Hb
end))
Anyhow, destruct on your assumption H will give you a proof for your goal if you are able to show exists x : X, ~ P x -> ⊥.
Instead of destruct, you could also do exfalso. apply H. to achieve the same thing.

Answer 2

Normally, destruct t applies when t is an inhabitant of an inductive type I, giving you one goal for each possible constructor for I that could have been used to produce t. Here as you remarked H has type P -> False, which is not an inductive type, but False is. So what happens is this: destruct gives you a first goal corresponding to the P hypothesis of H. Applying H to that goal leads to a term of type False, which is an inductive type, on which destruct works as it should, giving you zero goals since False has no constructors. Many tactics for inductive types work like this on hypothesis of the form P1 -> … -> Pn -> I where I is an inductive type: they give you side-goals for P1 … Pn, and then work on I.