Metamath Proof Explorer


Theorem ledivdiv

Description: Invert ratios of positive numbers and swap their ordering. (Contributed by NM, 9-Jan-2006)

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

Proof

Step Hyp Ref Expression
1 simpl ⊢ B ∈ ℝ ∧ 0 < B → B ∈ ℝ
2 gt0ne0 ⊢ B ∈ ℝ ∧ 0 < B → B ≠ 0
3 1 2 jca ⊢ B ∈ ℝ ∧ 0 < B → B ∈ ℝ ∧ B ≠ 0
4 redivcl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 → A B ∈ ℝ
5 4 3expb ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 → A B ∈ ℝ
6 3 5 sylan2 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ 0 < B → A B ∈ ℝ
7 6 adantlr ⊢ A ∈ ℝ ∧ 0 < A ∧ B ∈ ℝ ∧ 0 < B → A B ∈ ℝ
8 divgt0 ⊢ A ∈ ℝ ∧ 0 < A ∧ B ∈ ℝ ∧ 0 < B → 0 < A B
9 7 8 jca ⊢ A ∈ ℝ ∧ 0 < A ∧ B ∈ ℝ ∧ 0 < B → A B ∈ ℝ ∧ 0 < A B
10 simpl ⊢ D ∈ ℝ ∧ 0 < D → D ∈ ℝ
11 gt0ne0 ⊢ D ∈ ℝ ∧ 0 < D → D ≠ 0
12 10 11 jca ⊢ D ∈ ℝ ∧ 0 < D → D ∈ ℝ ∧ D ≠ 0
13 redivcl ⊢ C ∈ ℝ ∧ D ∈ ℝ ∧ D ≠ 0 → C D ∈ ℝ
14 13 3expb ⊢ C ∈ ℝ ∧ D ∈ ℝ ∧ D ≠ 0 → C D ∈ ℝ
15 12 14 sylan2 ⊢ C ∈ ℝ ∧ D ∈ ℝ ∧ 0 < D → C D ∈ ℝ
16 15 adantlr ⊢ C ∈ ℝ ∧ 0 < C ∧ D ∈ ℝ ∧ 0 < D → C D ∈ ℝ
17 divgt0 ⊢ C ∈ ℝ ∧ 0 < C ∧ D ∈ ℝ ∧ 0 < D → 0 < C D
18 16 17 jca ⊢ C ∈ ℝ ∧ 0 < C ∧ D ∈ ℝ ∧ 0 < D → C D ∈ ℝ ∧ 0 < C D
19 lerec ⊢ A B ∈ ℝ ∧ 0 < A B ∧ C D ∈ ℝ ∧ 0 < C D → A B ≤ C D ↔ 1 C D ≤ 1 A B
20 9 18 19 syl2an ⊢ A ∈ ℝ ∧ 0 < A ∧ B ∈ ℝ ∧ 0 < B ∧ C ∈ ℝ ∧ 0 < C ∧ D ∈ ℝ ∧ 0 < D → A B ≤ C D ↔ 1 C D ≤ 1 A B
21 recn ⊢ C ∈ ℝ → C ∈ ℂ
22 21 adantr ⊢ C ∈ ℝ ∧ 0 < C → C ∈ ℂ
23 gt0ne0 ⊢ C ∈ ℝ ∧ 0 < C → C ≠ 0
24 22 23 jca ⊢ C ∈ ℝ ∧ 0 < C → C ∈ ℂ ∧ C ≠ 0
25 recn ⊢ D ∈ ℝ → D ∈ ℂ
26 25 adantr ⊢ D ∈ ℝ ∧ 0 < D → D ∈ ℂ
27 26 11 jca ⊢ D ∈ ℝ ∧ 0 < D → D ∈ ℂ ∧ D ≠ 0
28 recdiv ⊢ C ∈ ℂ ∧ C ≠ 0 ∧ D ∈ ℂ ∧ D ≠ 0 → 1 C D = D C
29 24 27 28 syl2an ⊢ C ∈ ℝ ∧ 0 < C ∧ D ∈ ℝ ∧ 0 < D → 1 C D = D C
30 recn ⊢ A ∈ ℝ → A ∈ ℂ
31 30 adantr ⊢ A ∈ ℝ ∧ 0 < A → A ∈ ℂ
32 gt0ne0 ⊢ A ∈ ℝ ∧ 0 < A → A ≠ 0
33 31 32 jca ⊢ A ∈ ℝ ∧ 0 < A → A ∈ ℂ ∧ A ≠ 0
34 recn ⊢ B ∈ ℝ → B ∈ ℂ
35 34 adantr ⊢ B ∈ ℝ ∧ 0 < B → B ∈ ℂ
36 35 2 jca ⊢ B ∈ ℝ ∧ 0 < B → B ∈ ℂ ∧ B ≠ 0
37 recdiv ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ B ∈ ℂ ∧ B ≠ 0 → 1 A B = B A
38 33 36 37 syl2an ⊢ A ∈ ℝ ∧ 0 < A ∧ B ∈ ℝ ∧ 0 < B → 1 A B = B A
39 29 38 breqan12rd ⊢ A ∈ ℝ ∧ 0 < A ∧ B ∈ ℝ ∧ 0 < B ∧ C ∈ ℝ ∧ 0 < C ∧ D ∈ ℝ ∧ 0 < D → 1 C D ≤ 1 A B ↔ D C ≤ B A
40 20 39 bitrd ⊢ A ∈ ℝ ∧ 0 < A ∧ B ∈ ℝ ∧ 0 < B ∧ C ∈ ℝ ∧ 0 < C ∧ D ∈ ℝ ∧ 0 < D → A B ≤ C D ↔ D C ≤ B A