Metamath Proof Explorer


Theorem bj-idreseq

Description: Sufficient condition for the restricted identity relation to agree with equality. Note that the instance of bj-ideqg with _V substituted for V is a direct consequence of bj-idreseq . This is a strengthening of resieq which should be proved from it (note that currently, resieq relies on ideq ). Note that the intersection in the antecedent is not very meaningful, but is a device to prove versions with either class assumed to be a set. It could be enough to prove the version with a disjunctive antecedent: |- ( ( A e. C \/ B e. C ) -> ... ) . (Contributed by BJ, 25-Dec-2023)

Ref Expression
Assertion bj-idreseq ⊢ A ∩ B ∈ C → A I ↾ C B ↔ A = B

Proof

Step Hyp Ref Expression
1 bj-brresdm ⊢ A I ↾ C B → A ∈ C
2 relres ⊢ Rel ⁡ I ↾ C
3 2 brrelex2i ⊢ A I ↾ C B → B ∈ V
4 1 3 jca ⊢ A I ↾ C B → A ∈ C ∧ B ∈ V
5 4 adantl ⊢ A ∩ B ∈ C ∧ A I ↾ C B → A ∈ C ∧ B ∈ V
6 eqimss ⊢ A = B → A ⊆ B
7 dfss2 ⊢ A ⊆ B ↔ A ∩ B = A
8 6 7 sylib ⊢ A = B → A ∩ B = A
9 8 adantl ⊢ A ∩ B ∈ C ∧ A = B → A ∩ B = A
10 simpl ⊢ A ∩ B ∈ C ∧ A = B → A ∩ B ∈ C
11 9 10 eqeltrrd ⊢ A ∩ B ∈ C ∧ A = B → A ∈ C
12 eqimss2 ⊢ A = B → B ⊆ A
13 sseqin2 ⊢ B ⊆ A ↔ A ∩ B = B
14 12 13 sylib ⊢ A = B → A ∩ B = B
15 14 adantl ⊢ A ∩ B ∈ C ∧ A = B → A ∩ B = B
16 15 10 eqeltrrd ⊢ A ∩ B ∈ C ∧ A = B → B ∈ C
17 16 elexd ⊢ A ∩ B ∈ C ∧ A = B → B ∈ V
18 11 17 jca ⊢ A ∩ B ∈ C ∧ A = B → A ∈ C ∧ B ∈ V
19 brres ⊢ B ∈ V → A I ↾ C B ↔ A ∈ C ∧ A I B
20 19 adantl ⊢ A ∈ C ∧ B ∈ V → A I ↾ C B ↔ A ∈ C ∧ A I B
21 eqeq12 ⊢ x = A ∧ y = B → x = y ↔ A = B
22 df-id ⊢ I = x y | x = y
23 21 22 brabga ⊢ A ∈ C ∧ B ∈ V → A I B ↔ A = B
24 23 anbi2d ⊢ A ∈ C ∧ B ∈ V → A ∈ C ∧ A I B ↔ A ∈ C ∧ A = B
25 simp3 ⊢ A ∈ C ∧ B ∈ V ∧ A ∈ C ∧ A = B → A = B
26 25 3expib ⊢ A ∈ C ∧ B ∈ V → A ∈ C ∧ A = B → A = B
27 3simpb ⊢ A ∈ C ∧ B ∈ V ∧ A = B → A ∈ C ∧ A = B
28 27 3expia ⊢ A ∈ C ∧ B ∈ V → A = B → A ∈ C ∧ A = B
29 26 28 impbid ⊢ A ∈ C ∧ B ∈ V → A ∈ C ∧ A = B ↔ A = B
30 20 24 29 3bitrd ⊢ A ∈ C ∧ B ∈ V → A I ↾ C B ↔ A = B
31 5 18 30 pm5.21nd ⊢ A ∩ B ∈ C → A I ↾ C B ↔ A = B