Theorem bj-inex1gALT 37588

Description: Proof of inex1g 5287 from sepg 5258 to then allow proving inex1 5285 from it. That does not reduce the combined proof size of inex1 5285 and inex1g 5287. (Contributed by BJ, 14-Jul-2026.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
bj-inex1gALT	⊢ (𝐴 ∈ 𝑉 → (𝐴 ∩ 𝐵) ∈ V)

Proof of Theorem bj-inex1gALT

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		sepg 5258	. . 3 ⊢ (𝐴 ∈ 𝑉 → ∃𝑥∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)))
2		dfcleq 2755	. . . . 5 ⊢ (𝑥 = (𝐴 ∩ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝑥 ↔ 𝑦 ∈ (𝐴 ∩ 𝐵)))
3		elin 3920	. . . . . . . 8 ⊢ (𝑦 ∈ (𝐴 ∩ 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
4	3	a1i 11	. . . . . . 7 ⊢ (𝐴 ∈ 𝑉 → (𝑦 ∈ (𝐴 ∩ 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)))
5	4	bibi2d 345	. . . . . 6 ⊢ (𝐴 ∈ 𝑉 → ((𝑦 ∈ 𝑥 ↔ 𝑦 ∈ (𝐴 ∩ 𝐵)) ↔ (𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))))
6	5	albidv 1949	. . . . 5 ⊢ (𝐴 ∈ 𝑉 → (∀𝑦(𝑦 ∈ 𝑥 ↔ 𝑦 ∈ (𝐴 ∩ 𝐵)) ↔ ∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))))
7	2, 6	bitrid 286	. . . 4 ⊢ (𝐴 ∈ 𝑉 → (𝑥 = (𝐴 ∩ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))))
8	7	exbidv 1950	. . 3 ⊢ (𝐴 ∈ 𝑉 → (∃𝑥 𝑥 = (𝐴 ∩ 𝐵) ↔ ∃𝑥∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))))
9	1, 8	mpbird 260	. 2 ⊢ (𝐴 ∈ 𝑉 → ∃𝑥 𝑥 = (𝐴 ∩ 𝐵))
10		isset 3468	. 2 ⊢ ((𝐴 ∩ 𝐵) ∈ V ↔ ∃𝑥 𝑥 = (𝐴 ∩ 𝐵))
11	9, 10	sylibr 237	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∩ 𝐵) ∈ V)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1567   = wceq 1569  ∃wex 1808   ∈ wcel 2142  Vcvv 3454   ∩ cin 3903

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734  ax-sep 5256

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3456  df-in 3911

This theorem is used by: (None)