Metamath Proof Explorer


Theorem cnsubrglem

Description: Lemma for resubdrg and friends. (Contributed by Mario Carneiro, 4-Dec-2014)

Ref Expression
Hypotheses cnsubglem.1 ( 𝑥𝐴𝑥 ∈ ℂ )
cnsubglem.2 ( ( 𝑥𝐴𝑦𝐴 ) → ( 𝑥 + 𝑦 ) ∈ 𝐴 )
cnsubglem.3 ( 𝑥𝐴 → - 𝑥𝐴 )
cnsubrglem.4 1 ∈ 𝐴
cnsubrglem.5 ( ( 𝑥𝐴𝑦𝐴 ) → ( 𝑥 · 𝑦 ) ∈ 𝐴 )
Assertion cnsubrglem 𝐴 ∈ ( SubRing ‘ ℂfld )

Proof

Step Hyp Ref Expression
1 cnsubglem.1 ( 𝑥𝐴𝑥 ∈ ℂ )
2 cnsubglem.2 ( ( 𝑥𝐴𝑦𝐴 ) → ( 𝑥 + 𝑦 ) ∈ 𝐴 )
3 cnsubglem.3 ( 𝑥𝐴 → - 𝑥𝐴 )
4 cnsubrglem.4 1 ∈ 𝐴
5 cnsubrglem.5 ( ( 𝑥𝐴𝑦𝐴 ) → ( 𝑥 · 𝑦 ) ∈ 𝐴 )
6 1 2 3 4 cnsubglem 𝐴 ∈ ( SubGrp ‘ ℂfld )
7 5 rgen2 𝑥𝐴𝑦𝐴 ( 𝑥 · 𝑦 ) ∈ 𝐴
8 cnring fld ∈ Ring
9 cnfldbas ℂ = ( Base ‘ ℂfld )
10 cnfld1 1 = ( 1r ‘ ℂfld )
11 cnfldmul · = ( .r ‘ ℂfld )
12 9 10 11 issubrg2 ( ℂfld ∈ Ring → ( 𝐴 ∈ ( SubRing ‘ ℂfld ) ↔ ( 𝐴 ∈ ( SubGrp ‘ ℂfld ) ∧ 1 ∈ 𝐴 ∧ ∀ 𝑥𝐴𝑦𝐴 ( 𝑥 · 𝑦 ) ∈ 𝐴 ) ) )
13 8 12 ax-mp ( 𝐴 ∈ ( SubRing ‘ ℂfld ) ↔ ( 𝐴 ∈ ( SubGrp ‘ ℂfld ) ∧ 1 ∈ 𝐴 ∧ ∀ 𝑥𝐴𝑦𝐴 ( 𝑥 · 𝑦 ) ∈ 𝐴 ) )
14 6 4 7 13 mpbir3an 𝐴 ∈ ( SubRing ‘ ℂfld )