Metamath Proof Explorer


Theorem rereccl

Description: Closure law for reciprocal. (Contributed by NM, 30-Apr-2005) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion rereccl ⊢ A ∈ ℝ ∧ A ≠ 0 → 1 A ∈ ℝ

Proof

Step Hyp Ref Expression
1 ax-rrecex ⊢ A ∈ ℝ ∧ A ≠ 0 → ∃ x ∈ ℝ A ⁢ x = 1
2 eqcom ⊢ x = 1 A ↔ 1 A = x
3 1cnd ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → 1 ∈ ℂ
4 simpr ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → x ∈ ℝ
5 4 recnd ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → x ∈ ℂ
6 simpll ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → A ∈ ℝ
7 6 recnd ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → A ∈ ℂ
8 simplr ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → A ≠ 0
9 divmul ⊢ 1 ∈ ℂ ∧ x ∈ ℂ ∧ A ∈ ℂ ∧ A ≠ 0 → 1 A = x ↔ A ⁢ x = 1
10 3 5 7 8 9 syl112anc ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → 1 A = x ↔ A ⁢ x = 1
11 2 10 bitrid ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ → x = 1 A ↔ A ⁢ x = 1
12 11 rexbidva ⊢ A ∈ ℝ ∧ A ≠ 0 → ∃ x ∈ ℝ x = 1 A ↔ ∃ x ∈ ℝ A ⁢ x = 1
13 1 12 mpbird ⊢ A ∈ ℝ ∧ A ≠ 0 → ∃ x ∈ ℝ x = 1 A
14 risset ⊢ 1 A ∈ ℝ ↔ ∃ x ∈ ℝ x = 1 A
15 13 14 sylibr ⊢ A ∈ ℝ ∧ A ≠ 0 → 1 A ∈ ℝ