Metamath Proof Explorer


Theorem iscnrm3rlem4

Description: Lemma for iscnrm3rlem8 . Given two disjoint subsets S and T of the underlying set of a topology J , if N is a superset of ( ( ( clsJ )S ) \ T ) , then it is a superset of S . (Contributed by Zhi Wang, 5-Sep-2024)

Ref Expression
Hypotheses iscnrm3rlem4.1 ⊢ φ → J ∈ Top
iscnrm3rlem4.2 ⊢ φ → S ⊆ ⋃ J
iscnrm3rlem4.3 ⊢ φ → S ∩ T = ∅
iscnrm3rlem4.4 ⊢ φ → cls ⁡ J ⁡ S ∖ T ⊆ N
Assertion iscnrm3rlem4 ⊢ φ → S ⊆ N

Proof

Step Hyp Ref Expression
1 iscnrm3rlem4.1 ⊢ φ → J ∈ Top
2 iscnrm3rlem4.2 ⊢ φ → S ⊆ ⋃ J
3 iscnrm3rlem4.3 ⊢ φ → S ∩ T = ∅
4 iscnrm3rlem4.4 ⊢ φ → cls ⁡ J ⁡ S ∖ T ⊆ N
5 indifdi ⊢ S ∩ cls ⁡ J ⁡ S ∖ T = S ∩ cls ⁡ J ⁡ S ∖ S ∩ T
6 5 a1i ⊢ φ → S ∩ cls ⁡ J ⁡ S ∖ T = S ∩ cls ⁡ J ⁡ S ∖ S ∩ T
7 3 difeq2d ⊢ φ → S ∩ cls ⁡ J ⁡ S ∖ S ∩ T = S ∩ cls ⁡ J ⁡ S ∖ ∅
8 dif0 ⊢ S ∩ cls ⁡ J ⁡ S ∖ ∅ = S ∩ cls ⁡ J ⁡ S
9 7 8 eqtrdi ⊢ φ → S ∩ cls ⁡ J ⁡ S ∖ S ∩ T = S ∩ cls ⁡ J ⁡ S
10 eqid ⊢ ⋃ J = ⋃ J
11 10 sscls ⊢ J ∈ Top ∧ S ⊆ ⋃ J → S ⊆ cls ⁡ J ⁡ S
12 1 2 11 syl2anc ⊢ φ → S ⊆ cls ⁡ J ⁡ S
13 dfss2 ⊢ S ⊆ cls ⁡ J ⁡ S ↔ S ∩ cls ⁡ J ⁡ S = S
14 12 13 sylib ⊢ φ → S ∩ cls ⁡ J ⁡ S = S
15 6 9 14 3eqtrd ⊢ φ → S ∩ cls ⁡ J ⁡ S ∖ T = S
16 dfss2 ⊢ S ⊆ cls ⁡ J ⁡ S ∖ T ↔ S ∩ cls ⁡ J ⁡ S ∖ T = S
17 15 16 sylibr ⊢ φ → S ⊆ cls ⁡ J ⁡ S ∖ T
18 17 4 sstrd ⊢ φ → S ⊆ N