Metamath Proof Explorer


Theorem kur14lem10

Description: Lemma for kur14 . Discharge the set T . (Contributed by Mario Carneiro, 11-Feb-2015)

Ref Expression
Hypotheses kur14lem10.j ⊢ 𝐽 ∈ Top
kur14lem10.x ⊢ 𝑋 = ∪ 𝐽
kur14lem10.k ⊢ 𝐾 = ( cls ‘ 𝐽 )
kur14lem10.s ⊢ 𝑆 = ∩ { 𝑥 ∈ 𝒫 𝒫 𝑋 ∣ ( 𝐴 ∈ 𝑥 ∧ ∀ 𝑦 ∈ 𝑥 { ( 𝑋 ∖ 𝑦 ) , ( 𝐾 ‘ 𝑦 ) } ⊆ 𝑥 ) }
kur14lem10.a ⊢ 𝐴 ⊆ 𝑋
Assertion kur14lem10 ( 𝑆 ∈ Fin ∧ ( ♯ ‘ 𝑆 ) ≤ 1 4 )

Proof

Step Hyp Ref Expression
1 kur14lem10.j ⊢ 𝐽 ∈ Top
2 kur14lem10.x ⊢ 𝑋 = ∪ 𝐽
3 kur14lem10.k ⊢ 𝐾 = ( cls ‘ 𝐽 )
4 kur14lem10.s ⊢ 𝑆 = ∩ { 𝑥 ∈ 𝒫 𝒫 𝑋 ∣ ( 𝐴 ∈ 𝑥 ∧ ∀ 𝑦 ∈ 𝑥 { ( 𝑋 ∖ 𝑦 ) , ( 𝐾 ‘ 𝑦 ) } ⊆ 𝑥 ) }
5 kur14lem10.a ⊢ 𝐴 ⊆ 𝑋
6 eqid ⊢ ( int ‘ 𝐽 ) = ( int ‘ 𝐽 )
7 eqid ⊢ ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) ) = ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) )
8 eqid ⊢ ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) ) = ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) )
9 eqid ⊢ ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ 𝐴 ) ) = ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ 𝐴 ) )
10 eqid ⊢ ( ( ( { 𝐴 , ( 𝑋 ∖ 𝐴 ) , ( 𝐾 ‘ 𝐴 ) } ∪ { ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) ) , ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) ) , ( ( int ‘ 𝐽 ) ‘ 𝐴 ) } ) ∪ { ( 𝐾 ‘ ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) ) ) , ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ 𝐴 ) ) , ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ 𝐴 ) ) } ) ∪ ( { ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) ) ) , ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ 𝐴 ) ) ) , ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) ) ) ) } ∪ { ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) ) ) ) , ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ 𝐴 ) ) ) } ) ) = ( ( ( { 𝐴 , ( 𝑋 ∖ 𝐴 ) , ( 𝐾 ‘ 𝐴 ) } ∪ { ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) ) , ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) ) , ( ( int ‘ 𝐽 ) ‘ 𝐴 ) } ) ∪ { ( 𝐾 ‘ ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) ) ) , ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ 𝐴 ) ) , ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ 𝐴 ) ) } ) ∪ ( { ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) ) ) , ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ 𝐴 ) ) ) , ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( 𝑋 ∖ ( 𝐾 ‘ 𝐴 ) ) ) ) } ∪ { ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( 𝑋 ∖ 𝐴 ) ) ) ) , ( ( int ‘ 𝐽 ) ‘ ( 𝐾 ‘ ( ( int ‘ 𝐽 ) ‘ 𝐴 ) ) ) } ) )
11 1 2 3 6 5 7 8 9 10 4 kur14lem9 ⊢ ( 𝑆 ∈ Fin ∧ ( ♯ ‘ 𝑆 ) ≤ 1 4 )