Metamath Proof Explorer


Theorem ltdivmuld

Description: 'Less than' relationship between division and multiplication. (Contributed by Mario Carneiro, 28-May-2016)

Ref Expression
Hypotheses ltmul1d.1 ⊢ φ → A ∈ ℝ
ltmul1d.2 ⊢ φ → B ∈ ℝ
ltmul1d.3 ⊢ φ → C ∈ ℝ +
Assertion ltdivmuld ⊢ φ → A C < B ↔ A < C ⁢ B

Proof

Step Hyp Ref Expression
1 ltmul1d.1 ⊢ φ → A ∈ ℝ
2 ltmul1d.2 ⊢ φ → B ∈ ℝ
3 ltmul1d.3 ⊢ φ → C ∈ ℝ +
4 3 rpregt0d ⊢ φ → C ∈ ℝ ∧ 0 < C
5 ltdivmul ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A C < B ↔ A < C ⁢ B
6 1 2 4 5 syl3anc ⊢ φ → A C < B ↔ A < C ⁢ B