Metamath Proof Explorer


Theorem dyaddisj

Description: Two closed dyadic rational intervals are either in a subset relationship or are almost disjoint (the interiors are disjoint). (Contributed by Mario Carneiro, 26-Mar-2015)

Ref Expression
Hypothesis dyadmbl.1 F=x,y0x2yx+12y
Assertion dyaddisj AranFBranF.A.B.B.A.A.B=

Proof

Step Hyp Ref Expression
1 dyadmbl.1 F=x,y0x2yx+12y
2 1 dyadf F:×02
3 ffn F:×02FFn×0
4 ovelrn FFn×0AranFac0A=aFc
5 ovelrn FFn×0BranFbd0B=bFd
6 4 5 anbi12d FFn×0AranFBranFac0A=aFcbd0B=bFd
7 2 3 6 mp2b AranFBranFac0A=aFcbd0B=bFd
8 reeanv abc0A=aFcd0B=bFdac0A=aFcbd0B=bFd
9 7 8 bitr4i AranFBranFabc0A=aFcd0B=bFd
10 reeanv c0d0A=aFcB=bFdc0A=aFcd0B=bFd
11 nn0re c0c
12 11 ad2antrl abc0d0c
13 nn0re d0d
14 13 ad2antll abc0d0d
15 1 dyaddisjlem abc0d0cd.aFc.bFd.bFd.aFc.aFc.bFd=
16 ancom abba
17 ancom c0d0d0c0
18 16 17 anbi12i abc0d0bad0c0
19 1 dyaddisjlem bad0c0dc.bFd.aFc.aFc.bFd.bFd.aFc=
20 18 19 sylanb abc0d0dc.bFd.aFc.aFc.bFd.bFd.aFc=
21 orcom .bFd.aFc.aFc.bFd.aFc.bFd.bFd.aFc
22 incom .bFd.aFc=.aFc.bFd
23 22 eqeq1i .bFd.aFc=.aFc.bFd=
24 21 23 orbi12i .bFd.aFc.aFc.bFd.bFd.aFc=.aFc.bFd.bFd.aFc.aFc.bFd=
25 df-3or .bFd.aFc.aFc.bFd.bFd.aFc=.bFd.aFc.aFc.bFd.bFd.aFc=
26 df-3or .aFc.bFd.bFd.aFc.aFc.bFd=.aFc.bFd.bFd.aFc.aFc.bFd=
27 24 25 26 3bitr4i .bFd.aFc.aFc.bFd.bFd.aFc=.aFc.bFd.bFd.aFc.aFc.bFd=
28 20 27 sylib abc0d0dc.aFc.bFd.bFd.aFc.aFc.bFd=
29 12 14 15 28 lecasei abc0d0.aFc.bFd.bFd.aFc.aFc.bFd=
30 simpl A=aFcB=bFdA=aFc
31 30 fveq2d A=aFcB=bFd.A=.aFc
32 simpr A=aFcB=bFdB=bFd
33 32 fveq2d A=aFcB=bFd.B=.bFd
34 31 33 sseq12d A=aFcB=bFd.A.B.aFc.bFd
35 33 31 sseq12d A=aFcB=bFd.B.A.bFd.aFc
36 30 fveq2d A=aFcB=bFd.A=.aFc
37 32 fveq2d A=aFcB=bFd.B=.bFd
38 36 37 ineq12d A=aFcB=bFd.A.B=.aFc.bFd
39 38 eqeq1d A=aFcB=bFd.A.B=.aFc.bFd=
40 34 35 39 3orbi123d A=aFcB=bFd.A.B.B.A.A.B=.aFc.bFd.bFd.aFc.aFc.bFd=
41 29 40 syl5ibrcom abc0d0A=aFcB=bFd.A.B.B.A.A.B=
42 41 rexlimdvva abc0d0A=aFcB=bFd.A.B.B.A.A.B=
43 10 42 biimtrrid abc0A=aFcd0B=bFd.A.B.B.A.A.B=
44 43 rexlimivv abc0A=aFcd0B=bFd.A.B.B.A.A.B=
45 9 44 sylbi AranFBranF.A.B.B.A.A.B=