Proof (not-quite) irrelevance when when destructing coq equality - coq

Question

I have a dependent type which models a finite path in a transition system. The transition system has a function R that yields a proposition saying whether there's an edge between states s and s' with label a. The finite path type is:
Inductive FinPathTail (s : S i) :=
| FPTNil: FinPathTail s
| FPTCons (a : Act i) (s' : S i) : R i s a s' ->
FinPathTail s' -> FinPathTail s.
(The "tail" bit is because this actually models all but the head of a path starting at s).
I've defined a CoInductive type for a possibly infinite PathTail (I'll stick it at the bottom so as to get to the question faster) and I have a function, fpt_to_pt, to transform a FinPathTail into a PathTail. This should "obviously" be injective, so I wanted to prove a lemma of this form:
Lemma fpt_to_pt_inj {s : S i} (fpt fpt' : FinPathTail s)
: (forall s s' : S i, {s = s'} + {s <> s'}) ->
fpt_to_pt fpt = fpt_to_pt fpt' -> fpt = fpt'.
When trying to prove this by induction on fpt, I quickly get to the case where both sides are known to be conses. The goal ends up looking something like:
PTCons s a s' r (fpt_to_pt fpt) = PTCons s a2 s'2 r2 (fpt_to_pt fpt') ->
FPTCons s a s' r fpt = FPTCons s a2 s'2 r2 fpt'
that I'd like to decompose with the injection tactic. The result ends up like this:
existT (fun s'0 : S i => PathTail s'0) s' (fpt_to_pt fpt) =
existT (fun s'0 : S i => PathTail s'0) s'2 (fpt_to_pt fpt') ->
s' = s'2 -> a = a2 -> FPTCons s a s' r fpt = FPTCons s a2 s'2 r2 fpt'
and using the inversion_sigma tactic, I can transform it to:
B : s' = s'2
C : a = a2
A0 : s' = s'2
A1 : eq_rect s' (fun a : S i => PathTail a) (fpt_to_pt fpt) s'2 A0 = fpt_to_pt fpt'
I think I understand why I need decidability for the source domain, in order to use inj_pair2_eq_dec. What I don't understand is: what happened to r and r2? I understand that I don't have proof irrelevance, but doesn't that mean that they must have been equal in order for the conses to be equal? Or am I misunderstanding something fundamental about propositions?
PS: Here's the coinductive definition for PathTail:
CoInductive PathTail (s : S i) :=
| PTNil: PathTail s
| PTCons (a : Act i) (s' : S i) : R i s a s' -> PathTail s' -> PathTail s.

Answer 1

Apparently the injection tactic ignores equalities between proofs by default, but you can override this behavior with the Keep Proof Equalities flag:
Inductive foo : nat -> Prop :=
| Foo (n : nat) : foo n.
Inductive bar :=
| Bar (n : nat) : foo n -> bar.
Lemma test n nn m mm : Bar n nn = Bar m mm -> False.
Proof.
intros H. injection H. (* No equality generated. *)
Abort.
Set Keep Proof Equalities.
Lemma test n nn m mm : Bar n nn = Bar m mm -> False.
Proof.
intros H. injection H. (* Equality generated. *)
Abort.