Metamath Proof Explorer


Theorem ntrss

Description: Subset relationship for interior. (Contributed by NM, 3-Oct-2007)

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

Proof

Step Hyp Ref Expression
1 clscld.1 ⊢ X = ⋃ J
2 sscon ⊢ T ⊆ S → X ∖ S ⊆ X ∖ T
3 2 adantl ⊢ S ⊆ X ∧ T ⊆ S → X ∖ S ⊆ X ∖ T
4 difss ⊢ X ∖ T ⊆ X
5 3 4 jctil ⊢ S ⊆ X ∧ T ⊆ S → X ∖ T ⊆ X ∧ X ∖ S ⊆ X ∖ T
6 1 clsss ⊢ J ∈ Top ∧ X ∖ T ⊆ X ∧ X ∖ S ⊆ X ∖ T → cls ⁡ J ⁡ X ∖ S ⊆ cls ⁡ J ⁡ X ∖ T
7 6 3expb ⊢ J ∈ Top ∧ X ∖ T ⊆ X ∧ X ∖ S ⊆ X ∖ T → cls ⁡ J ⁡ X ∖ S ⊆ cls ⁡ J ⁡ X ∖ T
8 5 7 sylan2 ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → cls ⁡ J ⁡ X ∖ S ⊆ cls ⁡ J ⁡ X ∖ T
9 8 sscond ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → X ∖ cls ⁡ J ⁡ X ∖ T ⊆ X ∖ cls ⁡ J ⁡ X ∖ S
10 sstr2 ⊢ T ⊆ S → S ⊆ X → T ⊆ X
11 10 impcom ⊢ S ⊆ X ∧ T ⊆ S → T ⊆ X
12 1 ntrval2 ⊢ J ∈ Top ∧ T ⊆ X → int ⁡ J ⁡ T = X ∖ cls ⁡ J ⁡ X ∖ T
13 11 12 sylan2 ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → int ⁡ J ⁡ T = X ∖ cls ⁡ J ⁡ X ∖ T
14 1 ntrval2 ⊢ J ∈ Top ∧ S ⊆ X → int ⁡ J ⁡ S = X ∖ cls ⁡ J ⁡ X ∖ S
15 14 adantrr ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → int ⁡ J ⁡ S = X ∖ cls ⁡ J ⁡ X ∖ S
16 9 13 15 3sstr4d ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → int ⁡ J ⁡ T ⊆ int ⁡ J ⁡ S
17 16 3impb ⊢ J ∈ Top ∧ S ⊆ X ∧ T ⊆ S → int ⁡ J ⁡ T ⊆ int ⁡ J ⁡ S