Metamath Proof Explorer


Theorem reccl

Description: Closure law for reciprocal. (Contributed by NM, 30-Apr-2005)

Ref Expression
Assertion reccl ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 1 / 𝐴 ) ∈ ℂ )

Proof

Step Hyp Ref Expression
1 ax-1cn ⊢ 1 ∈ ℂ
2 divcl ⊢ ( ( 1 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 1 / 𝐴 ) ∈ ℂ )
3 1 2 mp3an1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 1 / 𝐴 ) ∈ ℂ )