Metamath Proof Explorer


Theorem efsubd

Description: Difference of exponents law for exponential function, deduction form. (Contributed by SN, 25-Apr-2025)

Ref Expression
Hypotheses efsubd.a ( 𝜑𝐴 ∈ ℂ )
efsubd.b ( 𝜑𝐵 ∈ ℂ )
Assertion efsubd ( 𝜑 → ( exp ‘ ( 𝐴𝐵 ) ) = ( ( exp ‘ 𝐴 ) / ( exp ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 efsubd.a ( 𝜑𝐴 ∈ ℂ )
2 efsubd.b ( 𝜑𝐵 ∈ ℂ )
3 efsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( exp ‘ ( 𝐴𝐵 ) ) = ( ( exp ‘ 𝐴 ) / ( exp ‘ 𝐵 ) ) )
4 1 2 3 syl2anc ( 𝜑 → ( exp ‘ ( 𝐴𝐵 ) ) = ( ( exp ‘ 𝐴 ) / ( exp ‘ 𝐵 ) ) )