Metamath Proof Explorer


Theorem cnsubrglem

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

Ref Expression
Hypotheses cnsubglem.1 x A x
cnsubglem.2 x A y A x + y A
cnsubglem.3 x A x A
cnsubrglem.4 1 A
cnsubrglem.5 x A y A x y A
Assertion cnsubrglem A SubRing fld

Proof

Step Hyp Ref Expression
1 cnsubglem.1 x A x
2 cnsubglem.2 x A y A x + y A
3 cnsubglem.3 x A x A
4 cnsubrglem.4 1 A
5 cnsubrglem.5 x A y A x y A
6 1 2 3 4 cnsubglem A SubGrp fld
7 5 rgen2 x A y A x y A
8 cnring fld Ring
9 cnfldbas = Base fld
10 cnfld1 1 = 1 fld
11 cnfldmul × = fld
12 9 10 11 issubrg2 fld Ring A SubRing fld A SubGrp fld 1 A x A y A x y A
13 8 12 ax-mp A SubRing fld A SubGrp fld 1 A x A y A x y A
14 6 4 7 13 mpbir3an A SubRing fld