Metamath Proof Explorer


Theorem dya2iocnrect

Description: For any point of an open rectangle in ( RR X. RR ) , there is a closed-below open-above dyadic rational square which contains that point and is included in the rectangle. (Contributed by Thierry Arnoux, 12-Oct-2017)

Ref Expression
Hypotheses sxbrsiga.0 J = topGen ran .
dya2ioc.1 I = x , n x 2 n x + 1 2 n
dya2ioc.2 R = u ran I , v ran I u × v
dya2iocnrect.1 B = ran e ran . , f ran . e × f
Assertion dya2iocnrect X 2 A B X A b ran R X b b A

Proof

Step Hyp Ref Expression
1 sxbrsiga.0 J = topGen ran .
2 dya2ioc.1 I = x , n x 2 n x + 1 2 n
3 dya2ioc.2 R = u ran I , v ran I u × v
4 dya2iocnrect.1 B = ran e ran . , f ran . e × f
5 4 eleq2i A B A ran e ran . , f ran . e × f
6 eqid e ran . , f ran . e × f = e ran . , f ran . e × f
7 vex e V
8 vex f V
9 7 8 xpex e × f V
10 6 9 elrnmpo A ran e ran . , f ran . e × f e ran . f ran . A = e × f
11 5 10 sylbb A B e ran . f ran . A = e × f
12 11 3ad2ant2 X 2 A B X A e ran . f ran . A = e × f
13 simp1 X 2 A B X A X 2
14 simp3 X 2 A B X A X A
15 12 13 14 jca32 X 2 A B X A e ran . f ran . A = e × f X 2 X A
16 r19.41vv e ran . f ran . A = e × f X 2 X A e ran . f ran . A = e × f X 2 X A
17 16 biimpri e ran . f ran . A = e × f X 2 X A e ran . f ran . A = e × f X 2 X A
18 simprl A = e × f X 2 X A X 2
19 simpl A = e × f X 2 X A A = e × f
20 simprr A = e × f X 2 X A X A
21 20 19 eleqtrd A = e × f X 2 X A X e × f
22 18 19 21 3jca A = e × f X 2 X A X 2 A = e × f X e × f
23 simpr e ran . f ran . X 2 A = e × f X e × f X 2 A = e × f X e × f
24 xp1st X 2 1 st X
25 24 3ad2ant1 X 2 A = e × f X e × f 1 st X
26 25 adantl e ran . f ran . X 2 A = e × f X e × f 1 st X
27 simpll e ran . f ran . X 2 A = e × f X e × f e ran .
28 xp1st X e × f 1 st X e
29 28 3ad2ant3 X 2 A = e × f X e × f 1 st X e
30 29 adantl e ran . f ran . X 2 A = e × f X e × f 1 st X e
31 1 2 dya2icoseg2 1 st X e ran . 1 st X e s ran I 1 st X s s e
32 26 27 30 31 syl3anc e ran . f ran . X 2 A = e × f X e × f s ran I 1 st X s s e
33 xp2nd X 2 2 nd X
34 33 3ad2ant1 X 2 A = e × f X e × f 2 nd X
35 34 adantl e ran . f ran . X 2 A = e × f X e × f 2 nd X
36 simplr e ran . f ran . X 2 A = e × f X e × f f ran .
37 xp2nd X e × f 2 nd X f
38 37 3ad2ant3 X 2 A = e × f X e × f 2 nd X f
39 38 adantl e ran . f ran . X 2 A = e × f X e × f 2 nd X f
40 1 2 dya2icoseg2 2 nd X f ran . 2 nd X f t ran I 2 nd X t t f
41 35 36 39 40 syl3anc e ran . f ran . X 2 A = e × f X e × f t ran I 2 nd X t t f
42 reeanv s ran I t ran I 1 st X s s e 2 nd X t t f s ran I 1 st X s s e t ran I 2 nd X t t f
43 32 41 42 sylanbrc e ran . f ran . X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f
44 eqid s × t = s × t
45 xpeq1 u = s u × v = s × v
46 45 eqeq2d u = s s × t = u × v s × t = s × v
47 xpeq2 v = t s × v = s × t
48 47 eqeq2d v = t s × t = s × v s × t = s × t
49 46 48 rspc2ev s ran I t ran I s × t = s × t u ran I v ran I s × t = u × v
50 44 49 mp3an3 s ran I t ran I u ran I v ran I s × t = u × v
51 vex u V
52 vex v V
53 51 52 xpex u × v V
54 3 53 elrnmpo s × t ran R u ran I v ran I s × t = u × v
55 50 54 sylibr s ran I t ran I s × t ran R
56 55 ad2antrl X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f s × t ran R
57 xpss 2 V × V
58 simpl1 X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f X 2
59 57 58 sselid X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f X V × V
60 simprrl X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f 1 st X s s e
61 60 simpld X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f 1 st X s
62 simprrr X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f 2 nd X t t f
63 62 simpld X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f 2 nd X t
64 elxp7 X s × t X V × V 1 st X s 2 nd X t
65 64 biimpri X V × V 1 st X s 2 nd X t X s × t
66 59 61 63 65 syl12anc X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f X s × t
67 60 simprd X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f s e
68 62 simprd X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f t f
69 xpss12 s e t f s × t e × f
70 67 68 69 syl2anc X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f s × t e × f
71 simpl2 X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f A = e × f
72 70 71 sseqtrrd X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f s × t A
73 eleq2 b = s × t X b X s × t
74 sseq1 b = s × t b A s × t A
75 73 74 anbi12d b = s × t X b b A X s × t s × t A
76 75 rspcev s × t ran R X s × t s × t A b ran R X b b A
77 56 66 72 76 syl12anc X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f b ran R X b b A
78 77 exp32 X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f b ran R X b b A
79 78 rexlimdvv X 2 A = e × f X e × f s ran I t ran I 1 st X s s e 2 nd X t t f b ran R X b b A
80 23 43 79 sylc e ran . f ran . X 2 A = e × f X e × f b ran R X b b A
81 22 80 sylan2 e ran . f ran . A = e × f X 2 X A b ran R X b b A
82 81 ex e ran . f ran . A = e × f X 2 X A b ran R X b b A
83 82 rexlimivv e ran . f ran . A = e × f X 2 X A b ran R X b b A
84 15 17 83 3syl X 2 A B X A b ran R X b b A