Metamath Proof Explorer


Theorem lt2subd

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

Ref Expression
Hypotheses leidd.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
ltnegd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
ltadd1d.3 ⊢ ( 𝜑 → 𝐶 ∈ ℝ )
lt2addd.4 ⊢ ( 𝜑 → 𝐷 ∈ ℝ )
lt2addd.5 ⊢ ( 𝜑 → 𝐴 < 𝐶 )
lt2addd.6 ⊢ ( 𝜑 → 𝐵 < 𝐷 )
Assertion lt2subd ( 𝜑 → ( 𝐴 − 𝐷 ) < ( 𝐶 − 𝐵 ) )

Proof

Step Hyp Ref Expression
1 leidd.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
2 ltnegd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
3 ltadd1d.3 ⊢ ( 𝜑 → 𝐶 ∈ ℝ )
4 lt2addd.4 ⊢ ( 𝜑 → 𝐷 ∈ ℝ )
5 lt2addd.5 ⊢ ( 𝜑 → 𝐴 < 𝐶 )
6 lt2addd.6 ⊢ ( 𝜑 → 𝐵 < 𝐷 )
7 lt2sub ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 𝐷 ∈ ℝ ) ∧ ( 𝐶 ∈ ℝ ∧ 𝐵 ∈ ℝ ) ) → ( ( 𝐴 < 𝐶 ∧ 𝐵 < 𝐷 ) → ( 𝐴 − 𝐷 ) < ( 𝐶 − 𝐵 ) ) )
8 1 4 3 2 7 syl22anc ⊢ ( 𝜑 → ( ( 𝐴 < 𝐶 ∧ 𝐵 < 𝐷 ) → ( 𝐴 − 𝐷 ) < ( 𝐶 − 𝐵 ) ) )
9 5 6 8 mp2and ⊢ ( 𝜑 → ( 𝐴 − 𝐷 ) < ( 𝐶 − 𝐵 ) )