Metamath Proof Explorer


Theorem abs3difi

Description: Absolute value of differences around common element. (Contributed by NM, 2-Oct-1999)

Ref Expression
Hypotheses absvalsqi.1 𝐴 ∈ ℂ
abssub.2 𝐵 ∈ ℂ
abs3dif.3 𝐶 ∈ ℂ
Assertion abs3difi ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 absvalsqi.1 𝐴 ∈ ℂ
2 abssub.2 𝐵 ∈ ℂ
3 abs3dif.3 𝐶 ∈ ℂ
4 abs3dif ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐵 ) ) ) )
5 1 2 3 4 mp3an ( abs ‘ ( 𝐴𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐵 ) ) )