Metamath Proof Explorer


Theorem omndmul3

Description: In an ordered monoid, the ordering is compatible with group power. This version does not require the monoid to be commutative. (Contributed by Thierry Arnoux, 23-Mar-2018)

Ref Expression
Hypotheses omndmul.0 B = Base M
omndmul.1 ˙ = M
omndmul3.m · ˙ = M
omndmul3.0 0 ˙ = 0 M
omndmul3.o φ M oMnd
omndmul3.1 φ N 0
omndmul3.2 φ P 0
omndmul3.3 φ N P
omndmul3.4 φ X B
omndmul3.5 φ 0 ˙ ˙ X
Assertion omndmul3 φ N · ˙ X ˙ P · ˙ X

Proof

Step Hyp Ref Expression
1 omndmul.0 B = Base M
2 omndmul.1 ˙ = M
3 omndmul3.m · ˙ = M
4 omndmul3.0 0 ˙ = 0 M
5 omndmul3.o φ M oMnd
6 omndmul3.1 φ N 0
7 omndmul3.2 φ P 0
8 omndmul3.3 φ N P
9 omndmul3.4 φ X B
10 omndmul3.5 φ 0 ˙ ˙ X
11 omndmnd M oMnd M Mnd
12 5 11 syl φ M Mnd
13 1 4 mndidcl M Mnd 0 ˙ B
14 12 13 syl φ 0 ˙ B
15 nn0sub N 0 P 0 N P P N 0
16 15 biimpa N 0 P 0 N P P N 0
17 6 7 8 16 syl21anc φ P N 0
18 1 3 mulgnn0cl M Mnd P N 0 X B P N · ˙ X B
19 12 17 9 18 syl3anc φ P N · ˙ X B
20 1 3 mulgnn0cl M Mnd N 0 X B N · ˙ X B
21 12 6 9 20 syl3anc φ N · ˙ X B
22 1 2 3 4 omndmul2 M oMnd X B P N 0 0 ˙ ˙ X 0 ˙ ˙ P N · ˙ X
23 5 9 17 10 22 syl121anc φ 0 ˙ ˙ P N · ˙ X
24 eqid + M = + M
25 1 2 24 omndadd M oMnd 0 ˙ B P N · ˙ X B N · ˙ X B 0 ˙ ˙ P N · ˙ X 0 ˙ + M N · ˙ X ˙ P N · ˙ X + M N · ˙ X
26 5 14 19 21 23 25 syl131anc φ 0 ˙ + M N · ˙ X ˙ P N · ˙ X + M N · ˙ X
27 1 24 4 mndlid M Mnd N · ˙ X B 0 ˙ + M N · ˙ X = N · ˙ X
28 12 21 27 syl2anc φ 0 ˙ + M N · ˙ X = N · ˙ X
29 1 3 24 mulgnn0dir M Mnd P N 0 N 0 X B P - N + N · ˙ X = P N · ˙ X + M N · ˙ X
30 12 17 6 9 29 syl13anc φ P - N + N · ˙ X = P N · ˙ X + M N · ˙ X
31 7 nn0cnd φ P
32 6 nn0cnd φ N
33 31 32 npcand φ P - N + N = P
34 33 oveq1d φ P - N + N · ˙ X = P · ˙ X
35 30 34 eqtr3d φ P N · ˙ X + M N · ˙ X = P · ˙ X
36 26 28 35 3brtr3d φ N · ˙ X ˙ P · ˙ X