Metamath Proof Explorer


Theorem lediv12a

Description: Comparison of ratio of two nonnegative numbers. (Contributed by NM, 31-Dec-2005)

Ref Expression
Assertion lediv12a A B 0 A A B C D 0 < C C D A D B C

Proof

Step Hyp Ref Expression
1 simplr C D 0 < C C D D
2 0re 0
3 ltletr 0 C D 0 < C C D 0 < D
4 2 3 mp3an1 C D 0 < C C D 0 < D
5 4 imp C D 0 < C C D 0 < D
6 5 gt0ne0d C D 0 < C C D D 0
7 1 6 rereccld C D 0 < C C D 1 D
8 gt0ne0 C 0 < C C 0
9 rereccl C C 0 1 C
10 8 9 syldan C 0 < C 1 C
11 10 ad2ant2r C D 0 < C C D 1 C
12 recgt0 D 0 < D 0 < 1 D
13 1 5 12 syl2anc C D 0 < C C D 0 < 1 D
14 ltle 0 1 D 0 < 1 D 0 1 D
15 2 7 14 sylancr C D 0 < C C D 0 < 1 D 0 1 D
16 13 15 mpd C D 0 < C C D 0 1 D
17 simprr C D 0 < C C D C D
18 id C 0 < C C 0 < C
19 18 ad2ant2r C D 0 < C C D C 0 < C
20 lerec C 0 < C D 0 < D C D 1 D 1 C
21 19 1 5 20 syl12anc C D 0 < C C D C D 1 D 1 C
22 17 21 mpbid C D 0 < C C D 1 D 1 C
23 16 22 jca C D 0 < C C D 0 1 D 1 D 1 C
24 7 11 23 jca31 C D 0 < C C D 1 D 1 C 0 1 D 1 D 1 C
25 simplll A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C A
26 simplrl A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C 0 A
27 simpllr A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C B
28 25 26 27 jca31 A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C A 0 A B
29 simprll A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C 1 D
30 simprrl A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C 0 1 D
31 29 30 jca A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C 1 D 0 1 D
32 simprlr A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C 1 C
33 28 31 32 jca32 A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C A 0 A B 1 D 0 1 D 1 C
34 simplrr A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C A B
35 simprrr A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C 1 D 1 C
36 34 35 jca A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C A B 1 D 1 C
37 lemul12a A 0 A B 1 D 0 1 D 1 C A B 1 D 1 C A 1 D B 1 C
38 33 36 37 sylc A B 0 A A B 1 D 1 C 0 1 D 1 D 1 C A 1 D B 1 C
39 24 38 sylan2 A B 0 A A B C D 0 < C C D A 1 D B 1 C
40 recn A A
41 40 adantr A C D 0 < C C D A
42 recn D D
43 42 ad2antlr C D 0 < C C D D
44 43 adantl A C D 0 < C C D D
45 6 adantl A C D 0 < C C D D 0
46 41 44 45 divrecd A C D 0 < C C D A D = A 1 D
47 46 ad4ant14 A B 0 A A B C D 0 < C C D A D = A 1 D
48 recn B B
49 48 adantr B C 0 < C B
50 recn C C
51 50 ad2antrl B C 0 < C C
52 8 adantl B C 0 < C C 0
53 49 51 52 divrecd B C 0 < C B C = B 1 C
54 53 adantrrr B C 0 < C C D B C = B 1 C
55 54 adantrlr B C D 0 < C C D B C = B 1 C
56 55 ad4ant24 A B 0 A A B C D 0 < C C D B C = B 1 C
57 39 47 56 3brtr4d A B 0 A A B C D 0 < C C D A D B C