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 ( 𝜑𝐴 ∈ ℝ )
ltnegd.2 ( 𝜑𝐵 ∈ ℝ )
ltadd1d.3 ( 𝜑𝐶 ∈ ℝ )
lt2addd.4 ( 𝜑𝐷 ∈ ℝ )
le2addd.5 ( 𝜑𝐴𝐶 )
le2addd.6 ( 𝜑𝐵𝐷 )
Assertion le2addd ( 𝜑 → ( 𝐴 + 𝐵 ) ≤ ( 𝐶 + 𝐷 ) )

Proof

Step Hyp Ref Expression
1 leidd.1 ( 𝜑𝐴 ∈ ℝ )
2 ltnegd.2 ( 𝜑𝐵 ∈ ℝ )
3 ltadd1d.3 ( 𝜑𝐶 ∈ ℝ )
4 lt2addd.4 ( 𝜑𝐷 ∈ ℝ )
5 le2addd.5 ( 𝜑𝐴𝐶 )
6 le2addd.6 ( 𝜑𝐵𝐷 )
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 ) )