Metamath Proof Explorer


Theorem abs2dif2d

Description: Difference of absolute values. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses abscld.1 ( 𝜑𝐴 ∈ ℂ )
abssubd.2 ( 𝜑𝐵 ∈ ℂ )
Assertion abs2dif2d ( 𝜑 → ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ 𝐴 ) + ( abs ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ( 𝜑𝐴 ∈ ℂ )
2 abssubd.2 ( 𝜑𝐵 ∈ ℂ )
3 abs2dif2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ 𝐴 ) + ( abs ‘ 𝐵 ) ) )
4 1 2 3 syl2anc ( 𝜑 → ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ 𝐴 ) + ( abs ‘ 𝐵 ) ) )