Metamath Proof Explorer


Theorem isfne4b

Description: A condition for a topology to be finer than another. (Contributed by Jeff Hankins, 28-Sep-2009) (Revised by Mario Carneiro, 11-Sep-2015)

Ref Expression
Hypotheses isfne.1 𝑋 = 𝐴
isfne.2 𝑌 = 𝐵
Assertion isfne4b ( 𝐵𝑉 → ( 𝐴 Fne 𝐵 ↔ ( 𝑋 = 𝑌 ∧ ( topGen ‘ 𝐴 ) ⊆ ( topGen ‘ 𝐵 ) ) ) )

Proof

Step Hyp Ref Expression
1 isfne.1 𝑋 = 𝐴
2 isfne.2 𝑌 = 𝐵
3 1 2 isfne4 ( 𝐴 Fne 𝐵 ↔ ( 𝑋 = 𝑌𝐴 ⊆ ( topGen ‘ 𝐵 ) ) )
4 simpr ( ( 𝐵𝑉𝑋 = 𝑌 ) → 𝑋 = 𝑌 )
5 4 1 2 3eqtr3g ( ( 𝐵𝑉𝑋 = 𝑌 ) → 𝐴 = 𝐵 )
6 uniexg ( 𝐵𝑉 𝐵 ∈ V )
7 6 adantr ( ( 𝐵𝑉𝑋 = 𝑌 ) → 𝐵 ∈ V )
8 5 7 eqeltrd ( ( 𝐵𝑉𝑋 = 𝑌 ) → 𝐴 ∈ V )
9 uniexb ( 𝐴 ∈ V ↔ 𝐴 ∈ V )
10 8 9 sylibr ( ( 𝐵𝑉𝑋 = 𝑌 ) → 𝐴 ∈ V )
11 simpl ( ( 𝐵𝑉𝑋 = 𝑌 ) → 𝐵𝑉 )
12 tgss3 ( ( 𝐴 ∈ V ∧ 𝐵𝑉 ) → ( ( topGen ‘ 𝐴 ) ⊆ ( topGen ‘ 𝐵 ) ↔ 𝐴 ⊆ ( topGen ‘ 𝐵 ) ) )
13 10 11 12 syl2anc ( ( 𝐵𝑉𝑋 = 𝑌 ) → ( ( topGen ‘ 𝐴 ) ⊆ ( topGen ‘ 𝐵 ) ↔ 𝐴 ⊆ ( topGen ‘ 𝐵 ) ) )
14 13 pm5.32da ( 𝐵𝑉 → ( ( 𝑋 = 𝑌 ∧ ( topGen ‘ 𝐴 ) ⊆ ( topGen ‘ 𝐵 ) ) ↔ ( 𝑋 = 𝑌𝐴 ⊆ ( topGen ‘ 𝐵 ) ) ) )
15 3 14 bitr4id ( 𝐵𝑉 → ( 𝐴 Fne 𝐵 ↔ ( 𝑋 = 𝑌 ∧ ( topGen ‘ 𝐴 ) ⊆ ( topGen ‘ 𝐵 ) ) ) )