Metamath Proof Explorer


Theorem ntrkbimka

Description: If the interiors of disjoint sets are disjoint, then the interior of the empty set is the empty set. (Contributed by RP, 14-Jun-2021)

Ref Expression
Assertion ntrkbimka ⊢ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ ∅ = ∅

Proof

Step Hyp Ref Expression
1 inidm ⊢ I ⁡ ∅ ∩ I ⁡ ∅ = I ⁡ ∅
2 0elpw ⊢ ∅ ∈ 𝒫 B
3 ineq1 ⊢ s = ∅ → s ∩ t = ∅ ∩ t
4 3 eqeq1d ⊢ s = ∅ → s ∩ t = ∅ ↔ ∅ ∩ t = ∅
5 fveq2 ⊢ s = ∅ → I ⁡ s = I ⁡ ∅
6 5 ineq1d ⊢ s = ∅ → I ⁡ s ∩ I ⁡ t = I ⁡ ∅ ∩ I ⁡ t
7 6 eqeq1d ⊢ s = ∅ → I ⁡ s ∩ I ⁡ t = ∅ ↔ I ⁡ ∅ ∩ I ⁡ t = ∅
8 4 7 imbi12d ⊢ s = ∅ → s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ ↔ ∅ ∩ t = ∅ → I ⁡ ∅ ∩ I ⁡ t = ∅
9 0in ⊢ ∅ ∩ t = ∅
10 pm5.5 ⊢ ∅ ∩ t = ∅ → ∅ ∩ t = ∅ → I ⁡ ∅ ∩ I ⁡ t = ∅ ↔ I ⁡ ∅ ∩ I ⁡ t = ∅
11 9 10 ax-mp ⊢ ∅ ∩ t = ∅ → I ⁡ ∅ ∩ I ⁡ t = ∅ ↔ I ⁡ ∅ ∩ I ⁡ t = ∅
12 8 11 bitrdi ⊢ s = ∅ → s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ ↔ I ⁡ ∅ ∩ I ⁡ t = ∅
13 fveq2 ⊢ t = ∅ → I ⁡ t = I ⁡ ∅
14 13 ineq2d ⊢ t = ∅ → I ⁡ ∅ ∩ I ⁡ t = I ⁡ ∅ ∩ I ⁡ ∅
15 14 eqeq1d ⊢ t = ∅ → I ⁡ ∅ ∩ I ⁡ t = ∅ ↔ I ⁡ ∅ ∩ I ⁡ ∅ = ∅
16 12 15 rspc2v ⊢ ∅ ∈ 𝒫 B ∧ ∅ ∈ 𝒫 B → ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ ∅ ∩ I ⁡ ∅ = ∅
17 2 2 16 mp2an ⊢ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ ∅ ∩ I ⁡ ∅ = ∅
18 1 17 eqtr3id ⊢ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ ∅ = ∅