Metamath Proof Explorer


Theorem le2addd

Description: Adding both side of two inequalities. (Contributed by Mario Carneiro, 27-May-2016) (Proof shortened by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses leidd.1 ⊢ φ → A ∈ ℝ
ltnegd.2 ⊢ φ → B ∈ ℝ
ltadd1d.3 ⊢ φ → C ∈ ℝ
lt2addd.4 ⊢ φ → D ∈ ℝ
le2addd.5 ⊢ φ → A ≤ C
le2addd.6 ⊢ φ → B ≤ D
Assertion le2addd ⊢ φ → A + B ≤ C + D

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 le2addd.5 ⊢ φ → A ≤ C
6 le2addd.6 ⊢ φ → B ≤ D
7 1 2 readdcld ⊢ φ → A + B ∈ ℝ
8 3 2 readdcld ⊢ φ → C + B ∈ ℝ
9 3 4 readdcld ⊢ φ → C + D ∈ ℝ
10 1 3 2 5 leadd1dd ⊢ φ → A + B ≤ C + B
11 2 4 3 6 leadd2dd ⊢ φ → C + B ≤ C + D
12 7 8 9 10 11 letrd ⊢ φ → A + B ≤ C + D