Metamath Proof Explorer


Theorem lediv12a

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

Ref Expression
Assertion lediv12a ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ 0 ≤ A ∧ A ≤ B ∧ C ∈ ℝ ∧ D ∈ ℝ ∧ 0 < C ∧ C ≤ D → A D ≤ B C

Proof

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