Metamath Proof Explorer


Theorem sgrpnmndex

Description: There is a semigroup which is not a monoid. (Contributed by AV, 29-Jan-2020)

Ref Expression
Assertion sgrpnmndex m Smgrp m Mnd

Proof

Step Hyp Ref Expression
1 prhash2ex 0 1 = 2
2 eqid 0 1 = 0 1
3 prex 0 1 V
4 eqeq1 x = u x = 0 u = 0
5 4 ifbid x = u if x = 0 0 1 = if u = 0 0 1
6 eqidd y = v if u = 0 0 1 = if u = 0 0 1
7 5 6 cbvmpov x 0 1 , y 0 1 if x = 0 0 1 = u 0 1 , v 0 1 if u = 0 0 1
8 7 opeq2i + ndx x 0 1 , y 0 1 if x = 0 0 1 = + ndx u 0 1 , v 0 1 if u = 0 0 1
9 8 preq2i Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 = Base ndx 0 1 + ndx u 0 1 , v 0 1 if u = 0 0 1
10 9 grpbase 0 1 V 0 1 = Base Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1
11 3 10 ax-mp 0 1 = Base Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1
12 11 eqcomi Base Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 = 0 1
13 3 3 mpoex u 0 1 , v 0 1 if u = 0 0 1 V
14 9 grpplusg u 0 1 , v 0 1 if u = 0 0 1 V u 0 1 , v 0 1 if u = 0 0 1 = + Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1
15 13 14 ax-mp u 0 1 , v 0 1 if u = 0 0 1 = + Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1
16 15 eqcomi + Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 = u 0 1 , v 0 1 if u = 0 0 1
17 2 12 16 sgrp2nmndlem4 0 1 = 2 Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 Smgrp
18 neleq1 m = Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 m Mnd Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 Mnd
19 18 adantl 0 1 = 2 m = Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 m Mnd Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 Mnd
20 2 12 16 sgrp2nmndlem5 0 1 = 2 Base ndx 0 1 + ndx x 0 1 , y 0 1 if x = 0 0 1 Mnd
21 17 19 20 rspcedvd 0 1 = 2 m Smgrp m Mnd
22 1 21 ax-mp m Smgrp m Mnd