Metamath Proof Explorer


Theorem dvrfval

Description: Division operation in a ring. (Contributed by Mario Carneiro, 2-Jul-2014) (Revised by Mario Carneiro, 2-Dec-2014) (Proof shortened by AV, 2-Mar-2024)

Ref Expression
Hypotheses dvrval.b B = Base R
dvrval.t · ˙ = R
dvrval.u U = Unit R
dvrval.i I = inv r R
dvrval.d × ˙ = / r R
Assertion dvrfval × ˙ = x B , y U x · ˙ I y

Proof

Step Hyp Ref Expression
1 dvrval.b B = Base R
2 dvrval.t · ˙ = R
3 dvrval.u U = Unit R
4 dvrval.i I = inv r R
5 dvrval.d × ˙ = / r R
6 fveq2 r = R Base r = Base R
7 6 1 eqtr4di r = R Base r = B
8 fveq2 r = R Unit r = Unit R
9 8 3 eqtr4di r = R Unit r = U
10 fveq2 r = R r = R
11 10 2 eqtr4di r = R r = · ˙
12 eqidd r = R x = x
13 fveq2 r = R inv r r = inv r R
14 13 4 eqtr4di r = R inv r r = I
15 14 fveq1d r = R inv r r y = I y
16 11 12 15 oveq123d r = R x r inv r r y = x · ˙ I y
17 7 9 16 mpoeq123dv r = R x Base r , y Unit r x r inv r r y = x B , y U x · ˙ I y
18 df-dvr / r = r V x Base r , y Unit r x r inv r r y
19 1 fvexi B V
20 3 fvexi U V
21 19 20 mpoex x B , y U x · ˙ I y V
22 17 18 21 fvmpt R V / r R = x B , y U x · ˙ I y
23 fvprc ¬ R V / r R =
24 fvprc ¬ R V Base R =
25 1 24 eqtrid ¬ R V B =
26 25 orcd ¬ R V B = U =
27 0mpo0 B = U = x B , y U x · ˙ I y =
28 26 27 syl ¬ R V x B , y U x · ˙ I y =
29 23 28 eqtr4d ¬ R V / r R = x B , y U x · ˙ I y
30 22 29 pm2.61i / r R = x B , y U x · ˙ I y
31 5 30 eqtri × ˙ = x B , y U x · ˙ I y