Metamath Proof Explorer


Theorem cjsubd

Description: Complex conjugate distributes over subtraction. (Contributed by Thierry Arnoux, 1-Jul-2025)

Ref Expression
Hypotheses cjsubd.1 ( 𝜑𝐴 ∈ ℂ )
cjsubd.2 ( 𝜑𝐵 ∈ ℂ )
Assertion cjsubd ( 𝜑 → ( ∗ ‘ ( 𝐴𝐵 ) ) = ( ( ∗ ‘ 𝐴 ) − ( ∗ ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 cjsubd.1 ( 𝜑𝐴 ∈ ℂ )
2 cjsubd.2 ( 𝜑𝐵 ∈ ℂ )
3 cjsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ∗ ‘ ( 𝐴𝐵 ) ) = ( ( ∗ ‘ 𝐴 ) − ( ∗ ‘ 𝐵 ) ) )
4 1 2 3 syl2anc ( 𝜑 → ( ∗ ‘ ( 𝐴𝐵 ) ) = ( ( ∗ ‘ 𝐴 ) − ( ∗ ‘ 𝐵 ) ) )