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Metamath Proof Explorer

Theorem abs2difabsd

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

Ref Expression
Hypotheses abscld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
abssubd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
Assertion abs2difabsd ( 𝜑 → ( abs ‘ ( ( abs ‘ 𝐴 ) − ( abs ‘ 𝐵 ) ) ) ≤ ( abs ‘ ( 𝐴 − 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 abssubd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 abs2difabs ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( abs ‘ ( ( abs ‘ 𝐴 ) − ( abs ‘ 𝐵 ) ) ) ≤ ( abs ‘ ( 𝐴 − 𝐵 ) ) )
4 1 2 3 syl2anc ⊢ ( 𝜑 → ( abs ‘ ( ( abs ‘ 𝐴 ) − ( abs ‘ 𝐵 ) ) ) ≤ ( abs ‘ ( 𝐴 − 𝐵 ) ) )