Metamath Proof Explorer


Theorem cnsrexpcl

Description: Exponentiation is closed in number rings. (Contributed by Stefan O'Rear, 30-Nov-2014)

Ref Expression
Hypotheses cnsrexpcl.s φ S SubRing fld
cnsrexpcl.x φ X S
cnsrexpcl.y φ Y 0
Assertion cnsrexpcl φ X Y S

Proof

Step Hyp Ref Expression
1 cnsrexpcl.s φ S SubRing fld
2 cnsrexpcl.x φ X S
3 cnsrexpcl.y φ Y 0
4 oveq2 a = 0 X a = X 0
5 4 eleq1d a = 0 X a S X 0 S
6 5 imbi2d a = 0 φ X a S φ X 0 S
7 oveq2 a = b X a = X b
8 7 eleq1d a = b X a S X b S
9 8 imbi2d a = b φ X a S φ X b S
10 oveq2 a = b + 1 X a = X b + 1
11 10 eleq1d a = b + 1 X a S X b + 1 S
12 11 imbi2d a = b + 1 φ X a S φ X b + 1 S
13 oveq2 a = Y X a = X Y
14 13 eleq1d a = Y X a S X Y S
15 14 imbi2d a = Y φ X a S φ X Y S
16 cnfldbas = Base fld
17 16 subrgss S SubRing fld S
18 1 17 syl φ S
19 18 2 sseldd φ X
20 19 exp0d φ X 0 = 1
21 cnfld1 1 = 1 fld
22 21 subrg1cl S SubRing fld 1 S
23 1 22 syl φ 1 S
24 20 23 eqeltrd φ X 0 S
25 19 3ad2ant2 b 0 φ X b S X
26 simp1 b 0 φ X b S b 0
27 25 26 expp1d b 0 φ X b S X b + 1 = X b X
28 1 3ad2ant2 b 0 φ X b S S SubRing fld
29 simp3 b 0 φ X b S X b S
30 2 3ad2ant2 b 0 φ X b S X S
31 cnfldmul × = fld
32 31 subrgmcl S SubRing fld X b S X S X b X S
33 28 29 30 32 syl3anc b 0 φ X b S X b X S
34 27 33 eqeltrd b 0 φ X b S X b + 1 S
35 34 3exp b 0 φ X b S X b + 1 S
36 35 a2d b 0 φ X b S φ X b + 1 S
37 6 9 12 15 24 36 nn0ind Y 0 φ X Y S
38 3 37 mpcom φ X Y S