Theorem bj-inexeqex 36633

Description: Lemma for bj-opelid 36635 (but not specific to the identity relation): if the intersection of two classes is a set and the two classes are equal, then both are sets (all three classes are equal, so they all belong to 𝑉, but it is more convenient to have V in the consequent for theorems using it). (Contributed by BJ, 27-Dec-2023.)

Assertion

Ref	Expression
bj-inexeqex	⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ V))

Proof of Theorem bj-inexeqex

Step	Hyp	Ref	Expression
1		eqimss 4038	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
2		df-ss 3964	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ 𝐵) = 𝐴)
3	1, 2	sylib 217	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐴)
4		eleq1 2817	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → ((𝐴 ∩ 𝐵) ∈ 𝑉 ↔ 𝐴 ∈ 𝑉))
5	4	biimpac 478	. . . 4 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ (𝐴 ∩ 𝐵) = 𝐴) → 𝐴 ∈ 𝑉)
6	3, 5	sylan2 592	. . 3 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐴 ∈ 𝑉)
7	6	elexd 3492	. 2 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐴 ∈ V)
8		eqimss2 4039	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
9		sseqin2 4215	. . . . 5 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐴 ∩ 𝐵) = 𝐵)
10	8, 9	sylib 217	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐵)
11		eleq1 2817	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = 𝐵 → ((𝐴 ∩ 𝐵) ∈ 𝑉 ↔ 𝐵 ∈ 𝑉))
12	11	biimpac 478	. . . 4 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ (𝐴 ∩ 𝐵) = 𝐵) → 𝐵 ∈ 𝑉)
13	10, 12	sylan2 592	. . 3 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐵 ∈ 𝑉)
14	13	elexd 3492	. 2 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐵 ∈ V)
15	7, 14	jca 511	1 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → (𝐴 ∈ V ∧ 𝐵 ∈ V))

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1534   ∈ wcel 2099  Vcvv 3471   ∩ cin 3946   ⊆ wss 3947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-rab 3430  df-v 3473  df-in 3954  df-ss 3964

This theorem is referenced by:  bj-elsn0  36634  bj-opelid  36635  bj-ideqgALT  36637