Metamath Proof Explorer


Theorem ltsub1d

Description: Subtraction from both sides of 'less than'. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses leidd.1 ( 𝜑𝐴 ∈ ℝ )
ltnegd.2 ( 𝜑𝐵 ∈ ℝ )
ltadd1d.3 ( 𝜑𝐶 ∈ ℝ )
Assertion ltsub1d ( 𝜑 → ( 𝐴 < 𝐵 ↔ ( 𝐴𝐶 ) < ( 𝐵𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 leidd.1 ( 𝜑𝐴 ∈ ℝ )
2 ltnegd.2 ( 𝜑𝐵 ∈ ℝ )
3 ltadd1d.3 ( 𝜑𝐶 ∈ ℝ )
4 ltsub1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐴𝐶 ) < ( 𝐵𝐶 ) ) )
5 1 2 3 4 syl3anc ( 𝜑 → ( 𝐴 < 𝐵 ↔ ( 𝐴𝐶 ) < ( 𝐵𝐶 ) ) )