Metamath Proof Explorer


Theorem redvr

Description: The division operation of the field of reals. (Contributed by Thierry Arnoux, 1-Nov-2017)

Ref Expression
Assertion redvr ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) → ( 𝐴 ( /r ‘ ℝfld ) 𝐵 ) = ( 𝐴 / 𝐵 ) )

Proof

Step Hyp Ref Expression
1 resubdrg ( ℝ ∈ ( SubRing ‘ ℂfld ) ∧ ℝfld ∈ DivRing )
2 1 simpli ℝ ∈ ( SubRing ‘ ℂfld )
3 simp1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) → 𝐴 ∈ ℝ )
4 3simpc ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) → ( 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) )
5 1 simpri fld ∈ DivRing
6 rebase ℝ = ( Base ‘ ℝfld )
7 eqid ( Unit ‘ ℝfld ) = ( Unit ‘ ℝfld )
8 re0g 0 = ( 0g ‘ ℝfld )
9 6 7 8 drngunit ( ℝfld ∈ DivRing → ( 𝐵 ∈ ( Unit ‘ ℝfld ) ↔ ( 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) ) )
10 5 9 ax-mp ( 𝐵 ∈ ( Unit ‘ ℝfld ) ↔ ( 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) )
11 4 10 sylibr ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) → 𝐵 ∈ ( Unit ‘ ℝfld ) )
12 df-refld fld = ( ℂflds ℝ )
13 cnflddiv / = ( /r ‘ ℂfld )
14 eqid ( /r ‘ ℝfld ) = ( /r ‘ ℝfld )
15 12 13 7 14 subrgdv ( ( ℝ ∈ ( SubRing ‘ ℂfld ) ∧ 𝐴 ∈ ℝ ∧ 𝐵 ∈ ( Unit ‘ ℝfld ) ) → ( 𝐴 / 𝐵 ) = ( 𝐴 ( /r ‘ ℝfld ) 𝐵 ) )
16 2 3 11 15 mp3an2i ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝐴 ( /r ‘ ℝfld ) 𝐵 ) )
17 16 eqcomd ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐵 ≠ 0 ) → ( 𝐴 ( /r ‘ ℝfld ) 𝐵 ) = ( 𝐴 / 𝐵 ) )