Metamath Proof Explorer


Theorem absmuld

Description: Absolute value distributes over multiplication. Proposition 10-3.7(f) of Gleason p. 133. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses abscld.1 ( 𝜑𝐴 ∈ ℂ )
abssubd.2 ( 𝜑𝐵 ∈ ℂ )
Assertion absmuld ( 𝜑 → ( abs ‘ ( 𝐴 · 𝐵 ) ) = ( ( abs ‘ 𝐴 ) · ( abs ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ( 𝜑𝐴 ∈ ℂ )
2 abssubd.2 ( 𝜑𝐵 ∈ ℂ )
3 absmul ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( abs ‘ ( 𝐴 · 𝐵 ) ) = ( ( abs ‘ 𝐴 ) · ( abs ‘ 𝐵 ) ) )
4 1 2 3 syl2anc ( 𝜑 → ( abs ‘ ( 𝐴 · 𝐵 ) ) = ( ( abs ‘ 𝐴 ) · ( abs ‘ 𝐵 ) ) )