Metamath Proof Explorer


Theorem absvalsqd

Description: Square of value of absolute value function. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypothesis abscld.1 ( 𝜑𝐴 ∈ ℂ )
Assertion absvalsqd ( 𝜑 → ( ( abs ‘ 𝐴 ) ↑ 2 ) = ( 𝐴 · ( ∗ ‘ 𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ( 𝜑𝐴 ∈ ℂ )
2 absvalsq ( 𝐴 ∈ ℂ → ( ( abs ‘ 𝐴 ) ↑ 2 ) = ( 𝐴 · ( ∗ ‘ 𝐴 ) ) )
3 1 2 syl ( 𝜑 → ( ( abs ‘ 𝐴 ) ↑ 2 ) = ( 𝐴 · ( ∗ ‘ 𝐴 ) ) )