Metamath Proof Explorer


Theorem clsss

Description: Subset relationship for closure. (Contributed by NM, 10-Feb-2007)

Ref Expression
Hypothesis clscld.1 ⊢ X = ⋃ J
Assertion clsss ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → cls ⁡ J ⁡ T ⊆ cls ⁡ J ⁡ S

Proof

Step Hyp Ref Expression
1 clscld.1 ⊢ X = ⋃ J
2 sstr2 ⊢ T ⊆ S → S ⊆ x → T ⊆ x
3 2 adantr ⊢ T ⊆ S ∧ x ∈ Clsd ⁡ J → S ⊆ x → T ⊆ x
4 3 ss2rabdv ⊢ T ⊆ S → x ∈ Clsd ⁡ J | S ⊆ x ⊆ x ∈ Clsd ⁡ J | T ⊆ x
5 intss ⊢ x ∈ Clsd ⁡ J | S ⊆ x ⊆ x ∈ Clsd ⁡ J | T ⊆ x → ⋂ x ∈ Clsd ⁡ J | T ⊆ x ⊆ ⋂ x ∈ Clsd ⁡ J | S ⊆ x
6 4 5 syl ⊢ T ⊆ S → ⋂ x ∈ Clsd ⁡ J | T ⊆ x ⊆ ⋂ x ∈ Clsd ⁡ J | S ⊆ x
7 6 3ad2ant3 ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → ⋂ x ∈ Clsd ⁡ J | T ⊆ x ⊆ ⋂ x ∈ Clsd ⁡ J | S ⊆ x
8 simp1 ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → J ∈ Top
9 sstr2 ⊢ T ⊆ S → S ⊆ X → T ⊆ X
10 9 impcom ⊢ S ⊆ X ∧ T ⊆ S → T ⊆ X
11 10 3adant1 ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → T ⊆ X
12 1 clsval ⊢ J ∈ Top ∧ T ⊆ X → cls ⁡ J ⁡ T = ⋂ x ∈ Clsd ⁡ J | T ⊆ x
13 8 11 12 syl2anc ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → cls ⁡ J ⁡ T = ⋂ x ∈ Clsd ⁡ J | T ⊆ x
14 1 clsval ⊢ J ∈ Top ∧ S ⊆ X → cls ⁡ J ⁡ S = ⋂ x ∈ Clsd ⁡ J | S ⊆ x
15 14 3adant3 ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → cls ⁡ J ⁡ S = ⋂ x ∈ Clsd ⁡ J | S ⊆ x
16 7 13 15 3sstr4d ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → cls ⁡ J ⁡ T ⊆ cls ⁡ J ⁡ S