Metamath Proof Explorer


Theorem fprodle

Description: If all the terms of two finite products are nonnegative and compare, so do the two products. (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses fprodle.kph ⊢ Ⅎ k φ
fprodle.a ⊢ φ → A ∈ Fin
fprodle.b ⊢ φ ∧ k ∈ A → B ∈ ℝ
fprodle.0l3b ⊢ φ ∧ k ∈ A → 0 ≤ B
fprodle.c ⊢ φ ∧ k ∈ A → C ∈ ℝ
fprodle.blec ⊢ φ ∧ k ∈ A → B ≤ C
Assertion fprodle ⊢ φ → ∏ k ∈ A B ≤ ∏ k ∈ A C

Proof

Step Hyp Ref Expression
1 fprodle.kph ⊢ Ⅎ k φ
2 fprodle.a ⊢ φ → A ∈ Fin
3 fprodle.b ⊢ φ ∧ k ∈ A → B ∈ ℝ
4 fprodle.0l3b ⊢ φ ∧ k ∈ A → 0 ≤ B
5 fprodle.c ⊢ φ ∧ k ∈ A → C ∈ ℝ
6 fprodle.blec ⊢ φ ∧ k ∈ A → B ≤ C
7 1red ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → 1 ∈ ℝ
8 nfra1 ⊢ Ⅎ k ∀ k ∈ A B ≠ 0
9 1 8 nfan ⊢ Ⅎ k φ ∧ ∀ k ∈ A B ≠ 0
10 2 adantr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → A ∈ Fin
11 5 adantlr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → C ∈ ℝ
12 3 adantlr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → B ∈ ℝ
13 rspa ⊢ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → B ≠ 0
14 13 adantll ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → B ≠ 0
15 11 12 14 redivcld ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → C B ∈ ℝ
16 9 10 15 fprodreclf ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A C B ∈ ℝ
17 1 2 3 fprodreclf ⊢ φ → ∏ k ∈ A B ∈ ℝ
18 17 adantr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ∈ ℝ
19 1 2 3 4 fprodge0 ⊢ φ → 0 ≤ ∏ k ∈ A B
20 19 adantr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → 0 ≤ ∏ k ∈ A B
21 4 adantlr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → 0 ≤ B
22 12 21 14 ne0gt0d ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → 0 < B
23 12 22 elrpd ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → B ∈ ℝ +
24 6 adantlr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → B ≤ C
25 divge1 ⊢ B ∈ ℝ + ∧ C ∈ ℝ ∧ B ≤ C → 1 ≤ C B
26 23 11 24 25 syl3anc ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → 1 ≤ C B
27 9 10 15 26 fprodge1 ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → 1 ≤ ∏ k ∈ A C B
28 7 16 18 20 27 lemul2ad ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ⋅ 1 ≤ ∏ k ∈ A B ⁢ ∏ k ∈ A C B
29 3 recnd ⊢ φ ∧ k ∈ A → B ∈ ℂ
30 1 2 29 fprodclf ⊢ φ → ∏ k ∈ A B ∈ ℂ
31 30 mulridd ⊢ φ → ∏ k ∈ A B ⋅ 1 = ∏ k ∈ A B
32 31 adantr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ⋅ 1 = ∏ k ∈ A B
33 5 recnd ⊢ φ ∧ k ∈ A → C ∈ ℂ
34 33 adantlr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → C ∈ ℂ
35 29 adantlr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 ∧ k ∈ A → B ∈ ℂ
36 9 10 34 35 14 fproddivf ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A C B = ∏ k ∈ A C ∏ k ∈ A B
37 36 oveq2d ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ⁢ ∏ k ∈ A C B = ∏ k ∈ A B ⁢ ∏ k ∈ A C ∏ k ∈ A B
38 1 2 33 fprodclf ⊢ φ → ∏ k ∈ A C ∈ ℂ
39 38 adantr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A C ∈ ℂ
40 30 adantr ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ∈ ℂ
41 9 10 35 14 fprodn0f ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ≠ 0
42 39 40 41 divcan2d ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ⁢ ∏ k ∈ A C ∏ k ∈ A B = ∏ k ∈ A C
43 37 42 eqtrd ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ⁢ ∏ k ∈ A C B = ∏ k ∈ A C
44 28 32 43 3brtr3d ⊢ φ ∧ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ≤ ∏ k ∈ A C
45 nne ⊢ ¬ B ≠ 0 ↔ B = 0
46 45 rexbii ⊢ ∃ k ∈ A ¬ B ≠ 0 ↔ ∃ k ∈ A B = 0
47 rexnal ⊢ ∃ k ∈ A ¬ B ≠ 0 ↔ ¬ ∀ k ∈ A B ≠ 0
48 nfv ⊢ Ⅎ j B = 0
49 nfcsb1v ⊢ Ⅎ _ k ⦋ j / k⦌ B
50 49 nfeq1 ⊢ Ⅎ k ⦋ j / k⦌ B = 0
51 csbeq1a ⊢ k = j → B = ⦋ j / k⦌ B
52 51 eqeq1d ⊢ k = j → B = 0 ↔ ⦋ j / k⦌ B = 0
53 48 50 52 cbvrexw ⊢ ∃ k ∈ A B = 0 ↔ ∃ j ∈ A ⦋ j / k⦌ B = 0
54 46 47 53 3bitr3i ⊢ ¬ ∀ k ∈ A B ≠ 0 ↔ ∃ j ∈ A ⦋ j / k⦌ B = 0
55 nfv ⊢ Ⅎ k j ∈ A
56 1 55 50 nf3an ⊢ Ⅎ k φ ∧ j ∈ A ∧ ⦋ j / k⦌ B = 0
57 2 3ad2ant1 ⊢ φ ∧ j ∈ A ∧ ⦋ j / k⦌ B = 0 → A ∈ Fin
58 29 3ad2antl1 ⊢ φ ∧ j ∈ A ∧ ⦋ j / k⦌ B = 0 ∧ k ∈ A → B ∈ ℂ
59 simp2 ⊢ φ ∧ j ∈ A ∧ ⦋ j / k⦌ B = 0 → j ∈ A
60 52 biimparc ⊢ ⦋ j / k⦌ B = 0 ∧ k = j → B = 0
61 60 3ad2antl3 ⊢ φ ∧ j ∈ A ∧ ⦋ j / k⦌ B = 0 ∧ k = j → B = 0
62 56 57 58 59 61 fprodeq0g ⊢ φ ∧ j ∈ A ∧ ⦋ j / k⦌ B = 0 → ∏ k ∈ A B = 0
63 62 rexlimdv3a ⊢ φ → ∃ j ∈ A ⦋ j / k⦌ B = 0 → ∏ k ∈ A B = 0
64 63 imp ⊢ φ ∧ ∃ j ∈ A ⦋ j / k⦌ B = 0 → ∏ k ∈ A B = 0
65 0red ⊢ φ ∧ k ∈ A → 0 ∈ ℝ
66 65 3 5 4 6 letrd ⊢ φ ∧ k ∈ A → 0 ≤ C
67 1 2 5 66 fprodge0 ⊢ φ → 0 ≤ ∏ k ∈ A C
68 67 adantr ⊢ φ ∧ ∃ j ∈ A ⦋ j / k⦌ B = 0 → 0 ≤ ∏ k ∈ A C
69 64 68 eqbrtrd ⊢ φ ∧ ∃ j ∈ A ⦋ j / k⦌ B = 0 → ∏ k ∈ A B ≤ ∏ k ∈ A C
70 54 69 sylan2b ⊢ φ ∧ ¬ ∀ k ∈ A B ≠ 0 → ∏ k ∈ A B ≤ ∏ k ∈ A C
71 44 70 pm2.61dan ⊢ φ → ∏ k ∈ A B ≤ ∏ k ∈ A C