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