Metamath Proof Explorer
Description: Closure law for reciprocal. (Contributed by Mario Carneiro, 27-May-2016)
|
|
Ref |
Expression |
|
Hypotheses |
redivcld.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
|
|
rereccld.2 |
⊢ ( 𝜑 → 𝐴 ≠ 0 ) |
|
Assertion |
rereccld |
⊢ ( 𝜑 → ( 1 / 𝐴 ) ∈ ℝ ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
redivcld.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
2 |
|
rereccld.2 |
⊢ ( 𝜑 → 𝐴 ≠ 0 ) |
3 |
|
rereccl |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐴 ≠ 0 ) → ( 1 / 𝐴 ) ∈ ℝ ) |
4 |
1 2 3
|
syl2anc |
⊢ ( 𝜑 → ( 1 / 𝐴 ) ∈ ℝ ) |