Metamath Proof Explorer


Theorem ntrclskb

Description: The interiors of disjoint sets are disjoint if and only if the closures of sets that span the base set also span the base set. (Contributed by RP, 10-Jun-2021)

Ref Expression
Hypotheses ntrcls.o O = i V k 𝒫 i 𝒫 i j 𝒫 i i k i j
ntrcls.d D = O B
ntrcls.r φ I D K
Assertion ntrclskb φ s 𝒫 B t 𝒫 B s t = I s I t = s 𝒫 B t 𝒫 B s t = B K s K t = B

Proof

Step Hyp Ref Expression
1 ntrcls.o O = i V k 𝒫 i 𝒫 i j 𝒫 i i k i j
2 ntrcls.d D = O B
3 ntrcls.r φ I D K
4 ineq1 s = a s t = a t
5 4 eqeq1d s = a s t = a t =
6 fveq2 s = a I s = I a
7 6 ineq1d s = a I s I t = I a I t
8 7 eqeq1d s = a I s I t = I a I t =
9 5 8 imbi12d s = a s t = I s I t = a t = I a I t =
10 ineq2 t = b a t = a b
11 10 eqeq1d t = b a t = a b =
12 fveq2 t = b I t = I b
13 12 ineq2d t = b I a I t = I a I b
14 13 eqeq1d t = b I a I t = I a I b =
15 11 14 imbi12d t = b a t = I a I t = a b = I a I b =
16 9 15 cbvral2vw s 𝒫 B t 𝒫 B s t = I s I t = a 𝒫 B b 𝒫 B a b = I a I b =
17 2 3 ntrclsrcomplex φ B s 𝒫 B
18 17 adantr φ s 𝒫 B B s 𝒫 B
19 2 3 ntrclsrcomplex φ B a 𝒫 B
20 19 adantr φ a 𝒫 B B a 𝒫 B
21 difeq2 s = B a B s = B B a
22 21 eqeq2d s = B a a = B s a = B B a
23 22 adantl φ a 𝒫 B s = B a a = B s a = B B a
24 elpwi a 𝒫 B a B
25 dfss4 a B B B a = a
26 24 25 sylib a 𝒫 B B B a = a
27 26 eqcomd a 𝒫 B a = B B a
28 27 adantl φ a 𝒫 B a = B B a
29 20 23 28 rspcedvd φ a 𝒫 B s 𝒫 B a = B s
30 simpl1 φ s 𝒫 B a = B s t 𝒫 B φ
31 2 3 ntrclsrcomplex φ B t 𝒫 B
32 30 31 syl φ s 𝒫 B a = B s t 𝒫 B B t 𝒫 B
33 2 3 ntrclsrcomplex φ B b 𝒫 B
34 33 adantr φ b 𝒫 B B b 𝒫 B
35 difeq2 t = B b B t = B B b
36 35 eqeq2d t = B b b = B t b = B B b
37 36 adantl φ b 𝒫 B t = B b b = B t b = B B b
38 elpwi b 𝒫 B b B
39 dfss4 b B B B b = b
40 38 39 sylib b 𝒫 B B B b = b
41 40 eqcomd b 𝒫 B b = B B b
42 41 adantl φ b 𝒫 B b = B B b
43 34 37 42 rspcedvd φ b 𝒫 B t 𝒫 B b = B t
44 43 3ad2antl1 φ s 𝒫 B a = B s b 𝒫 B t 𝒫 B b = B t
45 simp13 φ s 𝒫 B a = B s t 𝒫 B b = B t a = B s
46 ineq1 a = B s a b = B s b
47 46 eqeq1d a = B s a b = B s b =
48 fveq2 a = B s I a = I B s
49 48 ineq1d a = B s I a I b = I B s I b
50 49 eqeq1d a = B s I a I b = I B s I b =
51 47 50 imbi12d a = B s a b = I a I b = B s b = I B s I b =
52 45 51 syl φ s 𝒫 B a = B s t 𝒫 B b = B t a b = I a I b = B s b = I B s I b =
53 simp3 φ s 𝒫 B a = B s t 𝒫 B b = B t b = B t
54 ineq2 b = B t B s b = B s B t
55 54 eqeq1d b = B t B s b = B s B t =
56 fveq2 b = B t I b = I B t
57 56 ineq2d b = B t I B s I b = I B s I B t
58 57 eqeq1d b = B t I B s I b = I B s I B t =
59 55 58 imbi12d b = B t B s b = I B s I b = B s B t = I B s I B t =
60 53 59 syl φ s 𝒫 B a = B s t 𝒫 B b = B t B s b = I B s I b = B s B t = I B s I B t =
61 simp11 φ s 𝒫 B a = B s t 𝒫 B b = B t φ
62 simp12 φ s 𝒫 B a = B s t 𝒫 B b = B t s 𝒫 B
63 simp2 φ s 𝒫 B a = B s t 𝒫 B b = B t t 𝒫 B
64 simp2 φ s 𝒫 B t 𝒫 B s 𝒫 B
65 64 elpwid φ s 𝒫 B t 𝒫 B s B
66 simp3 φ s 𝒫 B t 𝒫 B t 𝒫 B
67 66 elpwid φ s 𝒫 B t 𝒫 B t B
68 65 67 unssd φ s 𝒫 B t 𝒫 B s t B
69 ssid B B
70 rcompleq s t B B B s t = B B s t = B B
71 68 69 70 sylancl φ s 𝒫 B t 𝒫 B s t = B B s t = B B
72 difundi B s t = B s B t
73 difid B B =
74 72 73 eqeq12i B s t = B B B s B t =
75 71 74 bitr2di φ s 𝒫 B t 𝒫 B B s B t = s t = B
76 1 2 3 ntrclsiex φ I 𝒫 B 𝒫 B
77 76 3ad2ant1 φ s 𝒫 B t 𝒫 B I 𝒫 B 𝒫 B
78 elmapi I 𝒫 B 𝒫 B I : 𝒫 B 𝒫 B
79 77 78 syl φ s 𝒫 B t 𝒫 B I : 𝒫 B 𝒫 B
80 2 3 ntrclsbex φ B V
81 80 3ad2ant1 φ s 𝒫 B t 𝒫 B B V
82 difssd φ s 𝒫 B t 𝒫 B B s B
83 81 82 sselpwd φ s 𝒫 B t 𝒫 B B s 𝒫 B
84 79 83 ffvelrnd φ s 𝒫 B t 𝒫 B I B s 𝒫 B
85 84 elpwid φ s 𝒫 B t 𝒫 B I B s B
86 ssinss1 I B s B I B s I B t B
87 85 86 syl φ s 𝒫 B t 𝒫 B I B s I B t B
88 0ss B
89 rcompleq I B s I B t B B I B s I B t = B I B s I B t = B
90 87 88 89 sylancl φ s 𝒫 B t 𝒫 B I B s I B t = B I B s I B t = B
91 difindi B I B s I B t = B I B s B I B t
92 dif0 B = B
93 91 92 eqeq12i B I B s I B t = B B I B s B I B t = B
94 90 93 bitrdi φ s 𝒫 B t 𝒫 B I B s I B t = B I B s B I B t = B
95 75 94 imbi12d φ s 𝒫 B t 𝒫 B B s B t = I B s I B t = s t = B B I B s B I B t = B
96 eqid D I = D I
97 eqid D I s = D I s
98 1 2 81 77 96 64 97 dssmapfv3d φ s 𝒫 B t 𝒫 B D I s = B I B s
99 eqid D I t = D I t
100 1 2 81 77 96 66 99 dssmapfv3d φ s 𝒫 B t 𝒫 B D I t = B I B t
101 98 100 uneq12d φ s 𝒫 B t 𝒫 B D I s D I t = B I B s B I B t
102 1 2 3 ntrclsfv1 φ D I = K
103 102 3ad2ant1 φ s 𝒫 B t 𝒫 B D I = K
104 fveq1 D I = K D I s = K s
105 fveq1 D I = K D I t = K t
106 104 105 uneq12d D I = K D I s D I t = K s K t
107 103 106 syl φ s 𝒫 B t 𝒫 B D I s D I t = K s K t
108 101 107 eqtr3d φ s 𝒫 B t 𝒫 B B I B s B I B t = K s K t
109 108 eqeq1d φ s 𝒫 B t 𝒫 B B I B s B I B t = B K s K t = B
110 109 imbi2d φ s 𝒫 B t 𝒫 B s t = B B I B s B I B t = B s t = B K s K t = B
111 95 110 bitrd φ s 𝒫 B t 𝒫 B B s B t = I B s I B t = s t = B K s K t = B
112 61 62 63 111 syl3anc φ s 𝒫 B a = B s t 𝒫 B b = B t B s B t = I B s I B t = s t = B K s K t = B
113 52 60 112 3bitrd φ s 𝒫 B a = B s t 𝒫 B b = B t a b = I a I b = s t = B K s K t = B
114 32 44 113 ralxfrd2 φ s 𝒫 B a = B s b 𝒫 B a b = I a I b = t 𝒫 B s t = B K s K t = B
115 18 29 114 ralxfrd2 φ a 𝒫 B b 𝒫 B a b = I a I b = s 𝒫 B t 𝒫 B s t = B K s K t = B
116 16 115 syl5bb φ s 𝒫 B t 𝒫 B s t = I s I t = s 𝒫 B t 𝒫 B s t = B K s K t = B