Metamath Proof Explorer


Theorem ltsub2

Description: Subtraction of both sides of 'less than'. (Contributed by NM, 29-Sep-2005) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion ltsub2 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐶𝐵 ) < ( 𝐶𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 lesub2 ( ( 𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵𝐴 ↔ ( 𝐶𝐴 ) ≤ ( 𝐶𝐵 ) ) )
2 1 3com12 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵𝐴 ↔ ( 𝐶𝐴 ) ≤ ( 𝐶𝐵 ) ) )
3 2 notbid ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ¬ 𝐵𝐴 ↔ ¬ ( 𝐶𝐴 ) ≤ ( 𝐶𝐵 ) ) )
4 simp1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐴 ∈ ℝ )
5 simp2 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐵 ∈ ℝ )
6 4 5 ltnled ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ¬ 𝐵𝐴 ) )
7 simp3 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐶 ∈ ℝ )
8 7 5 resubcld ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶𝐵 ) ∈ ℝ )
9 7 4 resubcld ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶𝐴 ) ∈ ℝ )
10 8 9 ltnled ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐶𝐵 ) < ( 𝐶𝐴 ) ↔ ¬ ( 𝐶𝐴 ) ≤ ( 𝐶𝐵 ) ) )
11 3 6 10 3bitr4d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐶𝐵 ) < ( 𝐶𝐴 ) ) )