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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 ) )