Metamath Proof Explorer


Theorem cnfldsub

Description: The subtraction operator in the field of complex numbers. (Contributed by Mario Carneiro, 15-Jun-2015)

Ref Expression
Assertion cnfldsub = - fld

Proof

Step Hyp Ref Expression
1 cnfldbas = Base fld
2 cnfldadd + = + fld
3 eqid inv g fld = inv g fld
4 eqid - fld = - fld
5 1 2 3 4 grpsubval x y x - fld y = x + inv g fld y
6 cnfldneg y inv g fld y = y
7 6 adantl x y inv g fld y = y
8 7 oveq2d x y x + inv g fld y = x + y
9 negsub x y x + y = x y
10 5 8 9 3eqtrrd x y x y = x - fld y
11 10 mpoeq3ia x , y x y = x , y x - fld y
12 subf : ×
13 ffn : × Fn ×
14 12 13 ax-mp Fn ×
15 fnov Fn × = x , y x y
16 14 15 mpbi = x , y x y
17 cnring fld Ring
18 ringgrp fld Ring fld Grp
19 17 18 ax-mp fld Grp
20 1 4 grpsubf fld Grp - fld : ×
21 ffn - fld : × - fld Fn ×
22 19 20 21 mp2b - fld Fn ×
23 fnov - fld Fn × - fld = x , y x - fld y
24 22 23 mpbi - fld = x , y x - fld y
25 11 16 24 3eqtr4i = - fld