Metamath Proof Explorer


Theorem omndadd2d

Description: In a commutative left ordered monoid, the ordering is compatible with monoid addition. Double addition version. (Contributed by Thierry Arnoux, 30-Jan-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
omndadd2d.c φMCMnd
Assertion omndadd2d 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 omndadd2d.c φMCMnd
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 MMndZBYBZ+˙YB
20 16 8 7 19 syl3anc φZ+˙YB
21 1 3 mndcl MMndZBWBZ+˙WB
22 16 8 5 21 syl3anc φZ+˙WB
23 18 20 22 3jca φX+˙YBZ+˙YBZ+˙WB
24 1 2 3 omndadd MoMndXBZBYBX˙ZX+˙Y˙Z+˙Y
25 4 6 8 7 9 24 syl131anc φX+˙Y˙Z+˙Y
26 1 2 3 omndadd MoMndYBWBZBY˙WY+˙Z˙W+˙Z
27 4 7 5 8 10 26 syl131anc φY+˙Z˙W+˙Z
28 1 3 cmncom MCMndYBZBY+˙Z=Z+˙Y
29 11 7 8 28 syl3anc φY+˙Z=Z+˙Y
30 1 3 cmncom MCMndWBZBW+˙Z=Z+˙W
31 11 5 8 30 syl3anc φW+˙Z=Z+˙W
32 27 29 31 3brtr3d φZ+˙Y˙Z+˙W
33 1 2 postr MPosetX+˙YBZ+˙YBZ+˙WBX+˙Y˙Z+˙YZ+˙Y˙Z+˙WX+˙Y˙Z+˙W
34 33 imp MPosetX+˙YBZ+˙YBZ+˙WBX+˙Y˙Z+˙YZ+˙Y˙Z+˙WX+˙Y˙Z+˙W
35 14 23 25 32 34 syl22anc φX+˙Y˙Z+˙W