Metamath Proof Explorer


Theorem imsubd

Description: Imaginary part distributes over subtraction. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses recld.1 ( 𝜑𝐴 ∈ ℂ )
readdd.2 ( 𝜑𝐵 ∈ ℂ )
Assertion imsubd ( 𝜑 → ( ℑ ‘ ( 𝐴𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) − ( ℑ ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 recld.1 ( 𝜑𝐴 ∈ ℂ )
2 readdd.2 ( 𝜑𝐵 ∈ ℂ )
3 imsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℑ ‘ ( 𝐴𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) − ( ℑ ‘ 𝐵 ) ) )
4 1 2 3 syl2anc ( 𝜑 → ( ℑ ‘ ( 𝐴𝐵 ) ) = ( ( ℑ ‘ 𝐴 ) − ( ℑ ‘ 𝐵 ) ) )