Metamath Proof Explorer


Theorem cndrng

Description: The complex numbers form a division ring. (Contributed by Stefan O'Rear, 27-Nov-2014)

Ref Expression
Assertion cndrng fld ∈ DivRing

Proof

Step Hyp Ref Expression
1 cnfldbas ℂ = ( Base ‘ ℂfld )
2 1 a1i ( ⊤ → ℂ = ( Base ‘ ℂfld ) )
3 cnfldmul · = ( .r ‘ ℂfld )
4 3 a1i ( ⊤ → · = ( .r ‘ ℂfld ) )
5 cnfld0 0 = ( 0g ‘ ℂfld )
6 5 a1i ( ⊤ → 0 = ( 0g ‘ ℂfld ) )
7 cnfld1 1 = ( 1r ‘ ℂfld )
8 7 a1i ( ⊤ → 1 = ( 1r ‘ ℂfld ) )
9 cnring fld ∈ Ring
10 9 a1i ( ⊤ → ℂfld ∈ Ring )
11 mulne0 ( ( ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ∧ ( 𝑦 ∈ ℂ ∧ 𝑦 ≠ 0 ) ) → ( 𝑥 · 𝑦 ) ≠ 0 )
12 11 3adant1 ( ( ⊤ ∧ ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ∧ ( 𝑦 ∈ ℂ ∧ 𝑦 ≠ 0 ) ) → ( 𝑥 · 𝑦 ) ≠ 0 )
13 ax-1ne0 1 ≠ 0
14 13 a1i ( ⊤ → 1 ≠ 0 )
15 reccl ( ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) → ( 1 / 𝑥 ) ∈ ℂ )
16 15 adantl ( ( ⊤ ∧ ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ) → ( 1 / 𝑥 ) ∈ ℂ )
17 recne0 ( ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) → ( 1 / 𝑥 ) ≠ 0 )
18 17 adantl ( ( ⊤ ∧ ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ) → ( 1 / 𝑥 ) ≠ 0 )
19 recid2 ( ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) → ( ( 1 / 𝑥 ) · 𝑥 ) = 1 )
20 19 adantl ( ( ⊤ ∧ ( 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) ) → ( ( 1 / 𝑥 ) · 𝑥 ) = 1 )
21 2 4 6 8 10 12 14 16 18 20 isdrngd ( ⊤ → ℂfld ∈ DivRing )
22 21 mptru fld ∈ DivRing