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 BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=I=

Proof

Step Hyp Ref Expression
1 pwidg BVB𝒫B
2 1 ad2antrr BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=B𝒫B
3 0elpw 𝒫B
4 3 a1i BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=𝒫B
5 simprr BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=s𝒫Bt𝒫Bst=IsIt=
6 ineq1 s=Bst=Bt
7 6 eqeq1d s=Bst=Bt=
8 fveq2 s=BIs=IB
9 8 ineq1d s=BIsIt=IBIt
10 9 eqeq1d s=BIsIt=IBIt=
11 7 10 imbi12d s=Bst=IsIt=Bt=IBIt=
12 ineq2 t=Bt=B
13 12 eqeq1d t=Bt=B=
14 fveq2 t=It=I
15 14 ineq2d t=IBIt=IBI
16 15 eqeq1d t=IBIt=IBI=
17 13 16 imbi12d t=Bt=IBIt=B=IBI=
18 in0 B=
19 pm5.5 B=B=IBI=IBI=
20 18 19 mp1i t=B=IBI=IBI=
21 17 20 bitrd t=Bt=IBIt=IBI=
22 11 21 rspc2va B𝒫B𝒫Bs𝒫Bt𝒫Bst=IsIt=IBI=
23 2 4 5 22 syl21anc BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=IBI=
24 23 ex BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=IBI=
25 elmapi I𝒫B𝒫BI:𝒫B𝒫B
26 25 adantl BVI𝒫B𝒫BI:𝒫B𝒫B
27 3 a1i BVI𝒫B𝒫B𝒫B
28 26 27 ffvelcdmd BVI𝒫B𝒫BI𝒫B
29 28 elpwid BVI𝒫B𝒫BIB
30 simpl IB=Bs𝒫Bt𝒫Bst=IsIt=IB=B
31 ineq1 IB=BIBI=BI
32 incom BI=IB
33 31 32 eqtrdi IB=BIBI=IB
34 33 eqeq1d IB=BIBI=IB=
35 34 biimpd IB=BIBI=IB=
36 reldisj IBIB=IBB
37 36 biimpd IBIB=IBB
38 difid BB=
39 38 sseq2i IBBI
40 ss0 II=
41 39 40 sylbi IBBI=
42 37 41 syl6com IB=IBI=
43 35 42 syl6com IBI=IB=BIBI=
44 43 com13 IBIB=BIBI=I=
45 29 30 44 syl2im BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=IBI=I=
46 24 45 mpdd BVI𝒫B𝒫BIB=Bs𝒫Bt𝒫Bst=IsIt=I=