Theorem bj-inexeqex 37120

Description: Lemma for bj-opelid 37122 (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 4067	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
2		dfss2 3994	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ 𝐵) = 𝐴)
3	1, 2	sylib 218	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐴)
4		eleq1 2832	. . . . 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 3512	. 2 ⊢ (((𝐴 ∩ 𝐵) ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐴 ∈ V)
8		eqimss2 4068	. . . . 5 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
9		sseqin2 4244	. . . . 5 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐴 ∩ 𝐵) = 𝐵)
10	8, 9	sylib 218	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐵)
11		eleq1 2832	. . . . 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 3512	. 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 1537   ∈ wcel 2108  Vcvv 3488   ∩ cin 3975   ⊆ wss 3976

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-in 3983  df-ss 3993

This theorem is referenced by:  bj-elsn0  37121  bj-opelid  37122  bj-ideqgALT  37124