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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 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
efsubd.b ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
Assertion efsubd ( 𝜑 → ( exp ‘ ( 𝐴 − 𝐵 ) ) = ( ( exp ‘ 𝐴 ) / ( exp ‘ 𝐵 ) ) )

Proof

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