Metamath Proof Explorer


Theorem restclsseplem

Description: Lemma for restclssep . (Contributed by Zhi Wang, 2-Sep-2024)

Ref Expression
Hypotheses restcls2.1 ⊢ ( 𝜑 → 𝐽 ∈ Top )
restcls2.2 ⊢ ( 𝜑 → 𝑋 = ∪ 𝐽 )
restcls2.3 ⊢ ( 𝜑 → 𝑌 ⊆ 𝑋 )
restcls2.4 ⊢ ( 𝜑 → 𝐾 = ( 𝐽 ↾t 𝑌 ) )
restcls2.5 ⊢ ( 𝜑 → 𝑆 ∈ ( Clsd ‘ 𝐾 ) )
restclsseplem.6 ⊢ ( 𝜑 → ( 𝑆 ∩ 𝑇 ) = ∅ )
restclsseplem.7 ⊢ ( 𝜑 → 𝑇 ⊆ 𝑌 )
Assertion restclsseplem ( 𝜑 → ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ 𝑇 ) = ∅ )

Proof

Step Hyp Ref Expression
1 restcls2.1 ⊢ ( 𝜑 → 𝐽 ∈ Top )
2 restcls2.2 ⊢ ( 𝜑 → 𝑋 = ∪ 𝐽 )
3 restcls2.3 ⊢ ( 𝜑 → 𝑌 ⊆ 𝑋 )
4 restcls2.4 ⊢ ( 𝜑 → 𝐾 = ( 𝐽 ↾t 𝑌 ) )
5 restcls2.5 ⊢ ( 𝜑 → 𝑆 ∈ ( Clsd ‘ 𝐾 ) )
6 restclsseplem.6 ⊢ ( 𝜑 → ( 𝑆 ∩ 𝑇 ) = ∅ )
7 restclsseplem.7 ⊢ ( 𝜑 → 𝑇 ⊆ 𝑌 )
8 1 2 3 4 5 restcls2 ⊢ ( 𝜑 → 𝑆 = ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ 𝑌 ) )
9 8 ineq1d ⊢ ( 𝜑 → ( 𝑆 ∩ 𝑇 ) = ( ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ 𝑌 ) ∩ 𝑇 ) )
10 inass ⊢ ( ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ 𝑌 ) ∩ 𝑇 ) = ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ ( 𝑌 ∩ 𝑇 ) )
11 9 10 eqtrdi ⊢ ( 𝜑 → ( 𝑆 ∩ 𝑇 ) = ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ ( 𝑌 ∩ 𝑇 ) ) )
12 sseqin2 ⊢ ( 𝑇 ⊆ 𝑌 ↔ ( 𝑌 ∩ 𝑇 ) = 𝑇 )
13 7 12 sylib ⊢ ( 𝜑 → ( 𝑌 ∩ 𝑇 ) = 𝑇 )
14 13 ineq2d ⊢ ( 𝜑 → ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ ( 𝑌 ∩ 𝑇 ) ) = ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ 𝑇 ) )
15 11 6 14 3eqtr3rd ⊢ ( 𝜑 → ( ( ( cls ‘ 𝐽 ) ‘ 𝑆 ) ∩ 𝑇 ) = ∅ )