Metamath Proof Explorer


Theorem ltsub1

Description: Subtraction from both sides of 'less than'. (Contributed by FL, 3-Jan-2008) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion ltsub1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐴𝐶 ) < ( 𝐵𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 lesub1 ( ( 𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵𝐴 ↔ ( 𝐵𝐶 ) ≤ ( 𝐴𝐶 ) ) )
2 1 3com12 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵𝐴 ↔ ( 𝐵𝐶 ) ≤ ( 𝐴𝐶 ) ) )
3 2 notbid ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ¬ 𝐵𝐴 ↔ ¬ ( 𝐵𝐶 ) ≤ ( 𝐴𝐶 ) ) )
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 4 7 resubcld ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴𝐶 ) ∈ ℝ )
9 5 7 resubcld ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵𝐶 ) ∈ ℝ )
10 8 9 ltnled ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴𝐶 ) < ( 𝐵𝐶 ) ↔ ¬ ( 𝐵𝐶 ) ≤ ( 𝐴𝐶 ) ) )
11 3 6 10 3bitr4d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐴𝐶 ) < ( 𝐵𝐶 ) ) )