Metamath Proof Explorer


Theorem absdivd

Description: Absolute value distributes over division. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses abscld.1 ( 𝜑𝐴 ∈ ℂ )
abssubd.2 ( 𝜑𝐵 ∈ ℂ )
absdivd.2 ( 𝜑𝐵 ≠ 0 )
Assertion absdivd ( 𝜑 → ( abs ‘ ( 𝐴 / 𝐵 ) ) = ( ( abs ‘ 𝐴 ) / ( abs ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1 ( 𝜑𝐴 ∈ ℂ )
2 abssubd.2 ( 𝜑𝐵 ∈ ℂ )
3 absdivd.2 ( 𝜑𝐵 ≠ 0 )
4 absdiv ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( abs ‘ ( 𝐴 / 𝐵 ) ) = ( ( abs ‘ 𝐴 ) / ( abs ‘ 𝐵 ) ) )
5 1 2 3 4 syl3anc ( 𝜑 → ( abs ‘ ( 𝐴 / 𝐵 ) ) = ( ( abs ‘ 𝐴 ) / ( abs ‘ 𝐵 ) ) )