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Metamath Proof Explorer

Theorem cxprecd

Description: Complex exponentiation of a reciprocal. (Contributed by Mario Carneiro, 30-May-2016)

Ref Expression
Hypotheses rpcxpcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ+ )
cxprecd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
Assertion cxprecd ( 𝜑 → ( ( 1 / 𝐴 ) ↑𝑐 𝐵 ) = ( 1 / ( 𝐴 ↑𝑐 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 rpcxpcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ+ )
2 cxprecd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 cxprec ⊢ ( ( 𝐴 ∈ ℝ+ ∧ 𝐵 ∈ ℂ ) → ( ( 1 / 𝐴 ) ↑𝑐 𝐵 ) = ( 1 / ( 𝐴 ↑𝑐 𝐵 ) ) )
4 1 2 3 syl2anc ⊢ ( 𝜑 → ( ( 1 / 𝐴 ) ↑𝑐 𝐵 ) = ( 1 / ( 𝐴 ↑𝑐 𝐵 ) ) )