Metamath Proof Explorer


Theorem ntrk0kbimka

Description: If the interiors of disjoint sets are disjoint and the interior of the base set is the base set, then the interior of the empty set is the empty set. Obsolete version of ntrkbimka . (Contributed by RP, 12-Jun-2021)

Ref Expression
Assertion ntrk0kbimka ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ ∅ = ∅

Proof

Step Hyp Ref Expression
1 pwidg ⊢ B ∈ V → B ∈ 𝒫 B
2 1 ad2antrr ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B ∧ I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → B ∈ 𝒫 B
3 0elpw ⊢ ∅ ∈ 𝒫 B
4 3 a1i ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B ∧ I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → ∅ ∈ 𝒫 B
5 simprr ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B ∧ I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅
6 ineq1 ⊢ s = B → s ∩ t = B ∩ t
7 6 eqeq1d ⊢ s = B → s ∩ t = ∅ ↔ B ∩ t = ∅
8 fveq2 ⊢ s = B → I ⁡ s = I ⁡ B
9 8 ineq1d ⊢ s = B → I ⁡ s ∩ I ⁡ t = I ⁡ B ∩ I ⁡ t
10 9 eqeq1d ⊢ s = B → I ⁡ s ∩ I ⁡ t = ∅ ↔ I ⁡ B ∩ I ⁡ t = ∅
11 7 10 imbi12d ⊢ s = B → s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ ↔ B ∩ t = ∅ → I ⁡ B ∩ I ⁡ t = ∅
12 ineq2 ⊢ t = ∅ → B ∩ t = B ∩ ∅
13 12 eqeq1d ⊢ t = ∅ → B ∩ t = ∅ ↔ B ∩ ∅ = ∅
14 fveq2 ⊢ t = ∅ → I ⁡ t = I ⁡ ∅
15 14 ineq2d ⊢ t = ∅ → I ⁡ B ∩ I ⁡ t = I ⁡ B ∩ I ⁡ ∅
16 15 eqeq1d ⊢ t = ∅ → I ⁡ B ∩ I ⁡ t = ∅ ↔ I ⁡ B ∩ I ⁡ ∅ = ∅
17 13 16 imbi12d ⊢ t = ∅ → B ∩ t = ∅ → I ⁡ B ∩ I ⁡ t = ∅ ↔ B ∩ ∅ = ∅ → I ⁡ B ∩ I ⁡ ∅ = ∅
18 in0 ⊢ B ∩ ∅ = ∅
19 pm5.5 ⊢ B ∩ ∅ = ∅ → B ∩ ∅ = ∅ → I ⁡ B ∩ I ⁡ ∅ = ∅ ↔ I ⁡ B ∩ I ⁡ ∅ = ∅
20 18 19 mp1i ⊢ t = ∅ → B ∩ ∅ = ∅ → I ⁡ B ∩ I ⁡ ∅ = ∅ ↔ I ⁡ B ∩ I ⁡ ∅ = ∅
21 17 20 bitrd ⊢ t = ∅ → B ∩ t = ∅ → I ⁡ B ∩ I ⁡ t = ∅ ↔ I ⁡ B ∩ I ⁡ ∅ = ∅
22 11 21 rspc2va ⊢ B ∈ 𝒫 B ∧ ∅ ∈ 𝒫 B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ B ∩ I ⁡ ∅ = ∅
23 2 4 5 22 syl21anc ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B ∧ I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ B ∩ I ⁡ ∅ = ∅
24 23 ex ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ B ∩ I ⁡ ∅ = ∅
25 elmapi ⊢ I ∈ 𝒫 B 𝒫 B → I : 𝒫 B ⟶ 𝒫 B
26 25 adantl ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → I : 𝒫 B ⟶ 𝒫 B
27 3 a1i ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → ∅ ∈ 𝒫 B
28 26 27 ffvelcdmd ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → I ⁡ ∅ ∈ 𝒫 B
29 28 elpwid ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → I ⁡ ∅ ⊆ B
30 simpl ⊢ I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ B = B
31 ineq1 ⊢ I ⁡ B = B → I ⁡ B ∩ I ⁡ ∅ = B ∩ I ⁡ ∅
32 incom ⊢ B ∩ I ⁡ ∅ = I ⁡ ∅ ∩ B
33 31 32 eqtrdi ⊢ I ⁡ B = B → I ⁡ B ∩ I ⁡ ∅ = I ⁡ ∅ ∩ B
34 33 eqeq1d ⊢ I ⁡ B = B → I ⁡ B ∩ I ⁡ ∅ = ∅ ↔ I ⁡ ∅ ∩ B = ∅
35 34 biimpd ⊢ I ⁡ B = B → I ⁡ B ∩ I ⁡ ∅ = ∅ → I ⁡ ∅ ∩ B = ∅
36 reldisj ⊢ I ⁡ ∅ ⊆ B → I ⁡ ∅ ∩ B = ∅ ↔ I ⁡ ∅ ⊆ B ∖ B
37 36 biimpd ⊢ I ⁡ ∅ ⊆ B → I ⁡ ∅ ∩ B = ∅ → I ⁡ ∅ ⊆ B ∖ B
38 difid ⊢ B ∖ B = ∅
39 38 sseq2i ⊢ I ⁡ ∅ ⊆ B ∖ B ↔ I ⁡ ∅ ⊆ ∅
40 ss0 ⊢ I ⁡ ∅ ⊆ ∅ → I ⁡ ∅ = ∅
41 39 40 sylbi ⊢ I ⁡ ∅ ⊆ B ∖ B → I ⁡ ∅ = ∅
42 37 41 syl6com ⊢ I ⁡ ∅ ∩ B = ∅ → I ⁡ ∅ ⊆ B → I ⁡ ∅ = ∅
43 35 42 syl6com ⊢ I ⁡ B ∩ I ⁡ ∅ = ∅ → I ⁡ B = B → I ⁡ ∅ ⊆ B → I ⁡ ∅ = ∅
44 43 com13 ⊢ I ⁡ ∅ ⊆ B → I ⁡ B = B → I ⁡ B ∩ I ⁡ ∅ = ∅ → I ⁡ ∅ = ∅
45 29 30 44 syl2im ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ B ∩ I ⁡ ∅ = ∅ → I ⁡ ∅ = ∅
46 24 45 mpdd ⊢ B ∈ V ∧ I ∈ 𝒫 B 𝒫 B → I ⁡ B = B ∧ ∀ s ∈ 𝒫 B ∀ t ∈ 𝒫 B s ∩ t = ∅ → I ⁡ s ∩ I ⁡ t = ∅ → I ⁡ ∅ = ∅