Proof with try congruence.
intros A B C D; destruct (next_in_dom_rawnext _ _ _ A B) as [s' [z' [F G]]].
simpl in F, C. cut (stk s2 = stk s'). intro Heq; rewrite Heq in *; clear Heq.
destruct (val2loc (env_get (stk s') x))...
destruct F as [H I]; destruct C as [J K].
assert (L : join_sub (hp s') (hp s1)).
destruct G as [x0 [G1 G2]]; exists (hp x0)...
assert (M : joins (hp s1) (hp s2)).
destruct D as [x0 [D1 D2]]; exists (hp x0)...
generalize (next_next_join_False _ _ (hp s2) _ _ _ _ H L J)... intro N.
apply N... apply join_sub_refl.
destruct G as [? [[G0 G1] G']]; destruct D as [? [[D0 D1] D']].
solve [rewrite G0, G1, D0; auto].
Qed.

Lemma join_stacks_eq : forall s0 s1 s : state,
join s0 s1 s -> stk s0=stk s1 /\ stk s0=stk s.