Metamath Proof Explorer


Theorem cnsubdrglem

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
cnsubrglem.6 x A x 0 1 x A
Assertion cnsubdrglem A SubRing fld fld 𝑠 A DivRing

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 cnsubrglem.6 x A x 0 1 x A
7 1 2 3 4 5 cnsubrglem A SubRing fld
8 cndrng fld DivRing
9 eqid fld 𝑠 A = fld 𝑠 A
10 cnfld0 0 = 0 fld
11 eqid inv r fld = inv r fld
12 9 10 11 issubdrg fld DivRing A SubRing fld fld 𝑠 A DivRing x A 0 inv r fld x A
13 8 7 12 mp2an fld 𝑠 A DivRing x A 0 inv r fld x A
14 cnring fld Ring
15 1 ssriv A
16 ssdif A A 0 0
17 15 16 ax-mp A 0 0
18 17 sseli x A 0 x 0
19 cnfldbas = Base fld
20 19 10 8 drngui 0 = Unit fld
21 cnflddiv ÷ = / r fld
22 cnfld1 1 = 1 fld
23 19 20 21 22 11 ringinvdv fld Ring x 0 inv r fld x = 1 x
24 14 18 23 sylancr x A 0 inv r fld x = 1 x
25 eldifsn x A 0 x A x 0
26 25 6 sylbi x A 0 1 x A
27 24 26 eqeltrd x A 0 inv r fld x A
28 13 27 mprgbir fld 𝑠 A DivRing
29 7 28 pm3.2i A SubRing fld fld 𝑠 A DivRing