Metamath Proof Explorer


Theorem orngmul

Description: In an ordered ring, the ordering is compatible with the ring multiplication operation. (Contributed by Thierry Arnoux, 20-Jan-2018)

Ref Expression
Hypotheses orngmul.0 B=BaseR
orngmul.1 ˙=R
orngmul.2 0˙=0R
orngmul.3 ·˙=R
Assertion orngmul RoRingXB0˙˙XYB0˙˙Y0˙˙X·˙Y

Proof

Step Hyp Ref Expression
1 orngmul.0 B=BaseR
2 orngmul.1 ˙=R
3 orngmul.2 0˙=0R
4 orngmul.3 ·˙=R
5 simp2r RoRingXB0˙˙XYB0˙˙Y0˙˙X
6 simp3r RoRingXB0˙˙XYB0˙˙Y0˙˙Y
7 simp2l RoRingXB0˙˙XYB0˙˙YXB
8 simp3l RoRingXB0˙˙XYB0˙˙YYB
9 1 3 4 2 isorng RoRingRRingRoGrpaBbB0˙˙a0˙˙b0˙˙a·˙b
10 9 simp3bi RoRingaBbB0˙˙a0˙˙b0˙˙a·˙b
11 10 3ad2ant1 RoRingXB0˙˙XYB0˙˙YaBbB0˙˙a0˙˙b0˙˙a·˙b
12 breq2 a=X0˙˙a0˙˙X
13 12 anbi1d a=X0˙˙a0˙˙b0˙˙X0˙˙b
14 oveq1 a=Xa·˙b=X·˙b
15 14 breq2d a=X0˙˙a·˙b0˙˙X·˙b
16 13 15 imbi12d a=X0˙˙a0˙˙b0˙˙a·˙b0˙˙X0˙˙b0˙˙X·˙b
17 breq2 b=Y0˙˙b0˙˙Y
18 17 anbi2d b=Y0˙˙X0˙˙b0˙˙X0˙˙Y
19 oveq2 b=YX·˙b=X·˙Y
20 19 breq2d b=Y0˙˙X·˙b0˙˙X·˙Y
21 18 20 imbi12d b=Y0˙˙X0˙˙b0˙˙X·˙b0˙˙X0˙˙Y0˙˙X·˙Y
22 16 21 rspc2va XBYBaBbB0˙˙a0˙˙b0˙˙a·˙b0˙˙X0˙˙Y0˙˙X·˙Y
23 7 8 11 22 syl21anc RoRingXB0˙˙XYB0˙˙Y0˙˙X0˙˙Y0˙˙X·˙Y
24 5 6 23 mp2and RoRingXB0˙˙XYB0˙˙Y0˙˙X·˙Y