Metamath Proof Explorer


Theorem divalglem9

Description: Lemma for divalg . (Contributed by Paul Chapman, 21-Mar-2011) (Revised by AV, 2-Oct-2020)

Ref Expression
Hypotheses divalglem8.1 N
divalglem8.2 D
divalglem8.3 D0
divalglem8.4 S=r0|DNr
divalglem9.5 R=supS<
Assertion divalglem9 ∃!xSx<D

Proof

Step Hyp Ref Expression
1 divalglem8.1 N
2 divalglem8.2 D
3 divalglem8.3 D0
4 divalglem8.4 S=r0|DNr
5 divalglem9.5 R=supS<
6 1 2 3 4 divalglem2 supS<S
7 5 6 eqeltri RS
8 1 2 3 4 5 divalglem5 0RR<D
9 8 simpri R<D
10 breq1 x=Rx<DR<D
11 10 rspcev RSR<DxSx<D
12 7 9 11 mp2an xSx<D
13 oveq2 r=xNr=Nx
14 13 breq2d r=xDNrDNx
15 14 4 elrab2 xSx0DNx
16 15 simplbi xSx0
17 16 nn0zd xSx
18 oveq2 r=yNr=Ny
19 18 breq2d r=yDNrDNy
20 19 4 elrab2 ySy0DNy
21 20 simplbi ySy0
22 21 nn0zd ySy
23 zsubcl NxNx
24 1 23 mpan xNx
25 zsubcl NyNy
26 1 25 mpan yNy
27 24 26 anim12i xyNxNy
28 17 22 27 syl2an xSySNxNy
29 15 simprbi xSDNx
30 20 simprbi ySDNy
31 29 30 anim12i xSySDNxDNy
32 dvds2sub DNxNyDNxDNyDN-x-Ny
33 2 32 mp3an1 NxNyDNxDNyDN-x-Ny
34 28 31 33 sylc xSySDN-x-Ny
35 zcn xx
36 zcn yy
37 1 zrei N
38 37 recni N
39 38 subidi NN=0
40 39 oveq1i N-N-xy=0xy
41 0cn 0
42 subsub2 0xy0xy=0+y-x
43 41 42 mp3an1 xy0xy=0+y-x
44 40 43 eqtrid xyN-N-xy=0+y-x
45 sub4 NNxyN-N-xy=N-x-Ny
46 38 38 45 mpanl12 xyN-N-xy=N-x-Ny
47 subcl yxyx
48 47 ancoms xyyx
49 48 addid2d xy0+y-x=yx
50 44 46 49 3eqtr3d xyN-x-Ny=yx
51 35 36 50 syl2an xyN-x-Ny=yx
52 17 22 51 syl2an xSySN-x-Ny=yx
53 52 breq2d xSySDN-x-NyDyx
54 34 53 mpbid xSySDyx
55 zsubcl yxyx
56 55 ancoms xyyx
57 absdvdsb DyxDyxDyx
58 2 56 57 sylancr xyDyxDyx
59 17 22 58 syl2an xSySDyxDyx
60 54 59 mpbid xSySDyx
61 nnabscl DD0D
62 2 3 61 mp2an D
63 62 nnzi D
64 divides DyxDyxkkD=yx
65 63 56 64 sylancr xyDyxkkD=yx
66 17 22 65 syl2an xSySDyxkkD=yx
67 60 66 mpbid xSySkkD=yx
68 67 adantr xSySx<Dy<DkkD=yx
69 1 2 3 4 divalglem8 xSySx<Dy<DkkD=yxx=y
70 69 rexlimdv xSySx<Dy<DkkD=yxx=y
71 68 70 mpd xSySx<Dy<Dx=y
72 71 ex xSySx<Dy<Dx=y
73 72 rgen2 xSySx<Dy<Dx=y
74 breq1 x=yx<Dy<D
75 74 reu4 ∃!xSx<DxSx<DxSySx<Dy<Dx=y
76 12 73 75 mpbir2an ∃!xSx<D