Metamath Proof Explorer


Theorem mulgt0b1d

Description: Biconditional, deductive form of mulgt0 . The second factor is positive iff the product is. (Contributed by SN, 26-Jun-2024)

Ref Expression
Hypotheses mulgt0b1d.a ⊢ φ → A ∈ ℝ
mulgt0b1d.b ⊢ φ → B ∈ ℝ
mulgt0b1d.1 ⊢ φ → 0 < A
Assertion mulgt0b1d ⊢ φ → 0 < B ↔ 0 < A ⁢ B

Proof

Step Hyp Ref Expression
1 mulgt0b1d.a ⊢ φ → A ∈ ℝ
2 mulgt0b1d.b ⊢ φ → B ∈ ℝ
3 mulgt0b1d.1 ⊢ φ → 0 < A
4 1 adantr ⊢ φ ∧ 0 < B → A ∈ ℝ
5 2 adantr ⊢ φ ∧ 0 < B → B ∈ ℝ
6 3 adantr ⊢ φ ∧ 0 < B → 0 < A
7 simpr ⊢ φ ∧ 0 < B → 0 < B
8 4 5 6 7 mulgt0d ⊢ φ ∧ 0 < B → 0 < A ⁢ B
9 8 ex ⊢ φ → 0 < B → 0 < A ⁢ B
10 1 adantr ⊢ φ ∧ A ⁢ B ⁢ 0 - ℝ 1 < 0 → A ∈ ℝ
11 1re ⊢ 1 ∈ ℝ
12 rernegcl ⊢ 1 ∈ ℝ → 0 - ℝ 1 ∈ ℝ
13 11 12 mp1i ⊢ φ → 0 - ℝ 1 ∈ ℝ
14 2 13 remulcld ⊢ φ → B ⁢ 0 - ℝ 1 ∈ ℝ
15 14 adantr ⊢ φ ∧ A ⁢ B ⁢ 0 - ℝ 1 < 0 → B ⁢ 0 - ℝ 1 ∈ ℝ
16 3 adantr ⊢ φ ∧ A ⁢ B ⁢ 0 - ℝ 1 < 0 → 0 < A
17 1 recnd ⊢ φ → A ∈ ℂ
18 2 recnd ⊢ φ → B ∈ ℂ
19 13 recnd ⊢ φ → 0 - ℝ 1 ∈ ℂ
20 17 18 19 mulassd ⊢ φ → A ⁢ B ⁢ 0 - ℝ 1 = A ⁢ B ⁢ 0 - ℝ 1
21 20 breq1d ⊢ φ → A ⁢ B ⁢ 0 - ℝ 1 < 0 ↔ A ⁢ B ⁢ 0 - ℝ 1 < 0
22 21 biimpa ⊢ φ ∧ A ⁢ B ⁢ 0 - ℝ 1 < 0 → A ⁢ B ⁢ 0 - ℝ 1 < 0
23 10 15 16 22 mulgt0con2d ⊢ φ ∧ A ⁢ B ⁢ 0 - ℝ 1 < 0 → B ⁢ 0 - ℝ 1 < 0
24 23 ex ⊢ φ → A ⁢ B ⁢ 0 - ℝ 1 < 0 → B ⁢ 0 - ℝ 1 < 0
25 1 2 remulcld ⊢ φ → A ⁢ B ∈ ℝ
26 relt0neg2 ⊢ A ⁢ B ∈ ℝ → 0 < A ⁢ B ↔ 0 - ℝ A ⁢ B < 0
27 25 26 syl ⊢ φ → 0 < A ⁢ B ↔ 0 - ℝ A ⁢ B < 0
28 1red ⊢ φ → 1 ∈ ℝ
29 25 28 remulneg2d ⊢ φ → A ⁢ B ⁢ 0 - ℝ 1 = 0 - ℝ A ⁢ B ⋅ 1
30 ax-1rid ⊢ A ⁢ B ∈ ℝ → A ⁢ B ⋅ 1 = A ⁢ B
31 25 30 syl ⊢ φ → A ⁢ B ⋅ 1 = A ⁢ B
32 31 oveq2d ⊢ φ → 0 - ℝ A ⁢ B ⋅ 1 = 0 - ℝ A ⁢ B
33 29 32 eqtrd ⊢ φ → A ⁢ B ⁢ 0 - ℝ 1 = 0 - ℝ A ⁢ B
34 33 breq1d ⊢ φ → A ⁢ B ⁢ 0 - ℝ 1 < 0 ↔ 0 - ℝ A ⁢ B < 0
35 27 34 bitr4d ⊢ φ → 0 < A ⁢ B ↔ A ⁢ B ⁢ 0 - ℝ 1 < 0
36 relt0neg2 ⊢ B ∈ ℝ → 0 < B ↔ 0 - ℝ B < 0
37 2 36 syl ⊢ φ → 0 < B ↔ 0 - ℝ B < 0
38 2 28 remulneg2d ⊢ φ → B ⁢ 0 - ℝ 1 = 0 - ℝ B ⋅ 1
39 ax-1rid ⊢ B ∈ ℝ → B ⋅ 1 = B
40 2 39 syl ⊢ φ → B ⋅ 1 = B
41 40 oveq2d ⊢ φ → 0 - ℝ B ⋅ 1 = 0 - ℝ B
42 38 41 eqtrd ⊢ φ → B ⁢ 0 - ℝ 1 = 0 - ℝ B
43 42 breq1d ⊢ φ → B ⁢ 0 - ℝ 1 < 0 ↔ 0 - ℝ B < 0
44 37 43 bitr4d ⊢ φ → 0 < B ↔ B ⁢ 0 - ℝ 1 < 0
45 24 35 44 3imtr4d ⊢ φ → 0 < A ⁢ B → 0 < B
46 9 45 impbid ⊢ φ → 0 < B ↔ 0 < A ⁢ B