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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 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
abssubd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
abs3difd.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
Assertion abs3difd ( 𝜑 → ( abs ‘ ( 𝐴 − 𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴 − 𝐶 ) ) + ( abs ‘ ( 𝐶 − 𝐵 ) ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 abssubd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 abs3difd.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4 abs3dif ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( abs ‘ ( 𝐴 − 𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴 − 𝐶 ) ) + ( abs ‘ ( 𝐶 − 𝐵 ) ) ) )
5 1 2 3 4 syl3anc ⊢ ( 𝜑 → ( abs ‘ ( 𝐴 − 𝐵 ) ) ≤ ( ( abs ‘ ( 𝐴 − 𝐶 ) ) + ( abs ‘ ( 𝐶 − 𝐵 ) ) ) )