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Metamath Proof Explorer

Theorem ltmuldiv

Description: 'Less than' relationship between division and multiplication. (Contributed by NM, 12-Oct-1999) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion ltmuldiv ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ⁢ C < B ↔ A < B C

Proof

Step Hyp Ref Expression
1 simp1 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ∈ ℝ
2 simp3l ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → C ∈ ℝ
3 1 2 remulcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ⁢ C ∈ ℝ
4 ltdiv1 ⊢ A ⁢ C ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ⁢ C < B ↔ A ⁢ C C < B C
5 3 4 syld3an1 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ⁢ C < B ↔ A ⁢ C C < B C
6 1 recnd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ∈ ℂ
7 2 recnd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → C ∈ ℂ
8 simp3r ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → 0 < C
9 8 gt0ne0d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → C ≠ 0
10 6 7 9 divcan4d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ⁢ C C = A
11 10 breq1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ⁢ C C < B C ↔ A < B C
12 5 11 bitrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ 0 < C → A ⁢ C < B ↔ A < B C