Metamath Proof Explorer


Theorem divalglem10

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

Ref Expression
Hypotheses divalglem8.1 N
divalglem8.2 D
divalglem8.3 D0
divalglem8.4 S=r0|DNr
Assertion divalglem10 ∃!rq0rr<DN=qD+r

Proof

Step Hyp Ref Expression
1 divalglem8.1 N
2 divalglem8.2 D
3 divalglem8.3 D0
4 divalglem8.4 S=r0|DNr
5 eqid supS<=supS<
6 1 2 3 4 5 divalglem9 ∃!xSx<D
7 elnn0z x0x0x
8 7 anbi2i x<Dx0x<Dx0x
9 an12 x<Dx0xxx<D0x
10 ancom x<D0x0xx<D
11 10 anbi2i xx<D0xx0xx<D
12 9 11 bitri x<Dx0xx0xx<D
13 8 12 bitri x<Dx0x0xx<D
14 13 anbi1i x<Dx0qN=qD+xx0xx<DqN=qD+x
15 anass x0xx<DqN=qD+xx0xx<DqN=qD+x
16 14 15 bitri x<Dx0qN=qD+xx0xx<DqN=qD+x
17 oveq2 r=xqD+r=qD+x
18 17 eqeq2d r=xN=qD+rN=qD+x
19 18 rexbidv r=xqN=qD+rqN=qD+x
20 1 2 3 4 divalglem4 S=r0|qN=qD+r
21 19 20 elrab2 xSx0qN=qD+x
22 21 anbi2i x<DxSx<Dx0qN=qD+x
23 ancom xSx<Dx<DxS
24 anass x<Dx0qN=qD+xx<Dx0qN=qD+x
25 22 23 24 3bitr4i xSx<Dx<Dx0qN=qD+x
26 df-3an 0xx<DN=qD+x0xx<DN=qD+x
27 26 rexbii q0xx<DN=qD+xq0xx<DN=qD+x
28 r19.42v q0xx<DN=qD+x0xx<DqN=qD+x
29 27 28 bitri q0xx<DN=qD+x0xx<DqN=qD+x
30 29 anbi2i xq0xx<DN=qD+xx0xx<DqN=qD+x
31 16 25 30 3bitr4i xSx<Dxq0xx<DN=qD+x
32 31 eubii ∃!xxSx<D∃!xxq0xx<DN=qD+x
33 df-reu ∃!xSx<D∃!xxSx<D
34 df-reu ∃!xq0xx<DN=qD+x∃!xxq0xx<DN=qD+x
35 32 33 34 3bitr4i ∃!xSx<D∃!xq0xx<DN=qD+x
36 6 35 mpbi ∃!xq0xx<DN=qD+x
37 breq2 x=r0x0r
38 breq1 x=rx<Dr<D
39 oveq2 x=rqD+x=qD+r
40 39 eqeq2d x=rN=qD+xN=qD+r
41 37 38 40 3anbi123d x=r0xx<DN=qD+x0rr<DN=qD+r
42 41 rexbidv x=rq0xx<DN=qD+xq0rr<DN=qD+r
43 42 cbvreuvw ∃!xq0xx<DN=qD+x∃!rq0rr<DN=qD+r
44 36 43 mpbi ∃!rq0rr<DN=qD+r