Theorem bj-inex1gALT 37669

Description: Proof of inex1g 5282 from sepg 5253 to then allow proving inex1 5280 from it. That does not reduce the combined proof size of inex1 5280 and inex1g 5282. (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 5253	. . 3 ⊢ (𝐴 ∈ 𝑉 → ∃𝑥∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)))
2		dfcleq 2753	. . . . 5 ⊢ (𝑥 = (𝐴 ∩ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝑥 ↔ 𝑦 ∈ (𝐴 ∩ 𝐵)))
3		elin 3915	. . . . . . . 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 1953	. . . . 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 1954	. . 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 3464	. 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 401  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2145  Vcvv 3450   ∩ cin 3898

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732  ax-sep 5251

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-in 3906

This theorem is used by: (None)