Metamath Proof Explorer


Theorem le2subd

Description: Subtracting both sides of two 'less than or equal to' 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 )
le2addd.5 ⊢ ( 𝜑 → 𝐴 ≤ 𝐶 )
le2addd.6 ⊢ ( 𝜑 → 𝐵 ≤ 𝐷 )
Assertion le2subd ( 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 le2addd.5 ⊢ ( 𝜑 → 𝐴 ≤ 𝐶 )
6 le2addd.6 ⊢ ( 𝜑 → 𝐵 ≤ 𝐷 )
7 le2sub ⊢ ( ( ( 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 ) )