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Metamath Proof Explorer

Theorem recdiv2

Description: Division into a reciprocal. (Contributed by NM, 19-Oct-2007)

Ref Expression
Assertion recdiv2 ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ B ∈ ℂ ∧ B ≠ 0 → 1 A B = 1 A ⁢ B

Proof

Step Hyp Ref Expression
1 ax-1cn ⊢ 1 ∈ ℂ
2 divdiv1 ⊢ 1 ∈ ℂ ∧ A ∈ ℂ ∧ A ≠ 0 ∧ B ∈ ℂ ∧ B ≠ 0 → 1 A B = 1 A ⁢ B
3 1 2 mp3an1 ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ B ∈ ℂ ∧ B ≠ 0 → 1 A B = 1 A ⁢ B