Metamath Proof Explorer


Theorem cxpsubd

Description: Exponent subtraction law for complex exponentiation. (Contributed by Mario Carneiro, 30-May-2016)

Ref Expression
Hypotheses cxp0d.1 ( 𝜑𝐴 ∈ ℂ )
cxpefd.2 ( 𝜑𝐴 ≠ 0 )
cxpefd.3 ( 𝜑𝐵 ∈ ℂ )
cxpaddd.4 ( 𝜑𝐶 ∈ ℂ )
Assertion cxpsubd ( 𝜑 → ( 𝐴𝑐 ( 𝐵𝐶 ) ) = ( ( 𝐴𝑐 𝐵 ) / ( 𝐴𝑐 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 cxp0d.1 ( 𝜑𝐴 ∈ ℂ )
2 cxpefd.2 ( 𝜑𝐴 ≠ 0 )
3 cxpefd.3 ( 𝜑𝐵 ∈ ℂ )
4 cxpaddd.4 ( 𝜑𝐶 ∈ ℂ )
5 cxpsub ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴𝑐 ( 𝐵𝐶 ) ) = ( ( 𝐴𝑐 𝐵 ) / ( 𝐴𝑐 𝐶 ) ) )
6 1 2 3 4 5 syl211anc ( 𝜑 → ( 𝐴𝑐 ( 𝐵𝐶 ) ) = ( ( 𝐴𝑐 𝐵 ) / ( 𝐴𝑐 𝐶 ) ) )