Metamath Proof Explorer


Theorem bj-inexeqex

Description: Lemma for bj-opelid (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)

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

Proof

Step Hyp Ref Expression
1 eqimss ⊢ ( 𝐴 = 𝐵 → 𝐴 ⊆ 𝐵 )
2 dfss2 ⊢ ( 𝐴 ⊆ 𝐵 ↔ ( 𝐴 ∩ 𝐵 ) = 𝐴 )
3 1 2 sylib ⊢ ( 𝐴 = 𝐵 → ( 𝐴 ∩ 𝐵 ) = 𝐴 )
4 eleq1 ⊢ ( ( 𝐴 ∩ 𝐵 ) = 𝐴 → ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ↔ 𝐴 ∈ 𝑉 ) )
5 4 biimpac ⊢ ( ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ∧ ( 𝐴 ∩ 𝐵 ) = 𝐴 ) → 𝐴 ∈ 𝑉 )
6 3 5 sylan2 ⊢ ( ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ∧ 𝐴 = 𝐵 ) → 𝐴 ∈ 𝑉 )
7 6 elexd ⊢ ( ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ∧ 𝐴 = 𝐵 ) → 𝐴 ∈ V )
8 eqimss2 ⊢ ( 𝐴 = 𝐵 → 𝐵 ⊆ 𝐴 )
9 sseqin2 ⊢ ( 𝐵 ⊆ 𝐴 ↔ ( 𝐴 ∩ 𝐵 ) = 𝐵 )
10 8 9 sylib ⊢ ( 𝐴 = 𝐵 → ( 𝐴 ∩ 𝐵 ) = 𝐵 )
11 eleq1 ⊢ ( ( 𝐴 ∩ 𝐵 ) = 𝐵 → ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ↔ 𝐵 ∈ 𝑉 ) )
12 11 biimpac ⊢ ( ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ∧ ( 𝐴 ∩ 𝐵 ) = 𝐵 ) → 𝐵 ∈ 𝑉 )
13 10 12 sylan2 ⊢ ( ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ∧ 𝐴 = 𝐵 ) → 𝐵 ∈ 𝑉 )
14 13 elexd ⊢ ( ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ∧ 𝐴 = 𝐵 ) → 𝐵 ∈ V )
15 7 14 jca ⊢ ( ( ( 𝐴 ∩ 𝐵 ) ∈ 𝑉 ∧ 𝐴 = 𝐵 ) → ( 𝐴 ∈ V ∧ 𝐵 ∈ V ) )