Metamath Proof Explorer


Theorem abs3difd

Description: Absolute value of differences around common element. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses abscld.1 ( 𝜑𝐴 ∈ ℂ )
abssubd.2 ( 𝜑𝐵 ∈ ℂ )
abs3difd.3 ( 𝜑𝐶 ∈ ℂ )
Assertion abs3difd ( 𝜑 → ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐵 ) ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ( 𝜑𝐴 ∈ ℂ )
2 abssubd.2 ( 𝜑𝐵 ∈ ℂ )
3 abs3difd.3 ( 𝜑𝐶 ∈ ℂ )
4 abs3dif ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐵 ) ) ) )
5 1 2 3 4 syl3anc ( 𝜑 → ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐵 ) ) ) )