Metamath Proof Explorer


Theorem cnflddiv

Description: The division operation in the field of complex numbers. (Contributed by Stefan O'Rear, 27-Nov-2014) (Revised by Mario Carneiro, 2-Dec-2014)

Ref Expression
Assertion cnflddiv ÷=/rfld

Proof

Step Hyp Ref Expression
1 cnring fldRing
2 cnfldbas =Basefld
3 cnfld0 0=0fld
4 cndrng fldDivRing
5 2 3 4 drngui 0=Unitfld
6 eqid /rfld=/rfld
7 cnfldmul ×=fld
8 2 5 6 7 dvrcan1 fldRingxy0x/rfldyy=x
9 1 8 mp3an1 xy0x/rfldyy=x
10 9 oveq1d xy0x/rfldyyy=xy
11 2 5 6 dvrcl fldRingxy0x/rfldy
12 1 11 mp3an1 xy0x/rfldy
13 simpr xy0y0
14 eldifsn y0yy0
15 13 14 sylib xy0yy0
16 15 simpld xy0y
17 15 simprd xy0y0
18 12 16 17 divcan4d xy0x/rfldyyy=x/rfldy
19 10 18 eqtr3d xy0xy=x/rfldy
20 simpl xy0x
21 divval xyy0xy=ιz|yz=x
22 20 16 17 21 syl3anc xy0xy=ιz|yz=x
23 19 22 eqtr3d xy0x/rfldy=ιz|yz=x
24 eqid invrfld=invrfld
25 2 7 5 24 6 dvrval xy0x/rfldy=xinvrfldy
26 23 25 eqtr3d xy0ιz|yz=x=xinvrfldy
27 26 mpoeq3ia x,y0ιz|yz=x=x,y0xinvrfldy
28 df-div ÷=x,y0ιz|yz=x
29 2 7 5 24 6 dvrfval /rfld=x,y0xinvrfldy
30 27 28 29 3eqtr4i ÷=/rfld