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Metamath Proof Explorer

Theorem absnegd

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

Ref Expression
Hypothesis abscld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
Assertion absnegd ( 𝜑 → ( abs ‘ - 𝐴 ) = ( abs ‘ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 absneg ⊢ ( 𝐴 ∈ ℂ → ( abs ‘ - 𝐴 ) = ( abs ‘ 𝐴 ) )
3 1 2 syl ⊢ ( 𝜑 → ( abs ‘ - 𝐴 ) = ( abs ‘ 𝐴 ) )