Metamath Proof Explorer


Theorem omndadd2rd

Description: In a left- and right- ordered monoid, the ordering is compatible with monoid addition. Double addition version. (Contributed by Thierry Arnoux, 2-May-2018)

Ref Expression
Hypotheses omndadd.0 B=BaseM
omndadd.1 ˙=M
omndadd.2 +˙=+M
omndadd2d.m φMoMnd
omndadd2d.w φWB
omndadd2d.x φXB
omndadd2d.y φYB
omndadd2d.z φZB
omndadd2d.1 φX˙Z
omndadd2d.2 φY˙W
omndadd2rd.c φopp𝑔MoMnd
Assertion omndadd2rd φX+˙Y˙Z+˙W

Proof

Step Hyp Ref Expression
1 omndadd.0 B=BaseM
2 omndadd.1 ˙=M
3 omndadd.2 +˙=+M
4 omndadd2d.m φMoMnd
5 omndadd2d.w φWB
6 omndadd2d.x φXB
7 omndadd2d.y φYB
8 omndadd2d.z φZB
9 omndadd2d.1 φX˙Z
10 omndadd2d.2 φY˙W
11 omndadd2rd.c φopp𝑔MoMnd
12 omndtos MoMndMToset
13 tospos MTosetMPoset
14 4 12 13 3syl φMPoset
15 omndmnd MoMndMMnd
16 4 15 syl φMMnd
17 1 3 mndcl MMndXBYBX+˙YB
18 16 6 7 17 syl3anc φX+˙YB
19 1 3 mndcl MMndXBWBX+˙WB
20 16 6 5 19 syl3anc φX+˙WB
21 1 3 mndcl MMndZBWBZ+˙WB
22 16 8 5 21 syl3anc φZ+˙WB
23 18 20 22 3jca φX+˙YBX+˙WBZ+˙WB
24 1 2 3 omndaddr opp𝑔MoMndYBWBXBY˙WX+˙Y˙X+˙W
25 11 7 5 6 10 24 syl131anc φX+˙Y˙X+˙W
26 1 2 3 omndadd MoMndXBZBWBX˙ZX+˙W˙Z+˙W
27 4 6 8 5 9 26 syl131anc φX+˙W˙Z+˙W
28 1 2 postr MPosetX+˙YBX+˙WBZ+˙WBX+˙Y˙X+˙WX+˙W˙Z+˙WX+˙Y˙Z+˙W
29 28 imp MPosetX+˙YBX+˙WBZ+˙WBX+˙Y˙X+˙WX+˙W˙Z+˙WX+˙Y˙Z+˙W
30 14 23 25 27 29 syl22anc φX+˙Y˙Z+˙W