Theorem bj-inexeqex 37361

Description: Lemma for bj-opelid 37363 (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 3993	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
2		dfss2 3920	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ 𝐵) = 𝐴)
3	1, 2	sylib 218	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐴)
4		eleq1 2825	. . . . 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 594	. . 3 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐴 ∈ 𝑉)
7	6	elexd 3465	. 2 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐴 ∈ V)
8		eqimss2 3994	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
9		sseqin2 4176	. . . . 5 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐴 ∩ 𝐵) = 𝐵)
10	8, 9	sylib 218	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐵)
11		eleq1 2825	. . . . 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 594	. . 3 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐵 ∈ 𝑉)
14	13	elexd 3465	. 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 1542   ∈ wcel 2114  Vcvv 3441   ∩ cin 3901   ⊆ wss 3902

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3401  df-v 3443  df-in 3909  df-ss 3919

This theorem is referenced by:  bj-elsn0  37362  bj-opelid  37363  bj-ideqgALT  37365