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 syl5bb ⊢ ( ( ( 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 )