Metamath Proof Explorer


Theorem rereccld

Description: Closure law for reciprocal. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses redivcld.1 ( 𝜑𝐴 ∈ ℝ )
rereccld.2 ( 𝜑𝐴 ≠ 0 )
Assertion rereccld ( 𝜑 → ( 1 / 𝐴 ) ∈ ℝ )

Proof

Step Hyp Ref Expression
1 redivcld.1 ( 𝜑𝐴 ∈ ℝ )
2 rereccld.2 ( 𝜑𝐴 ≠ 0 )
3 rereccl ( ( 𝐴 ∈ ℝ ∧ 𝐴 ≠ 0 ) → ( 1 / 𝐴 ) ∈ ℝ )
4 1 2 3 syl2anc ( 𝜑 → ( 1 / 𝐴 ) ∈ ℝ )