Metamath Proof Explorer


Theorem mgm2nsgrplem4

Description: Lemma 4 for mgm2nsgrp : M is not a semigroup. (Contributed by AV, 28-Jan-2020) (Proof shortened by AV, 31-Jan-2020)

Ref Expression
Hypotheses mgm2nsgrp.s S = A B
mgm2nsgrp.b Base M = S
mgm2nsgrp.o + M = x S , y S if x = A y = A B A
Assertion mgm2nsgrplem4 S = 2 M Smgrp

Proof

Step Hyp Ref Expression
1 mgm2nsgrp.s S = A B
2 mgm2nsgrp.b Base M = S
3 mgm2nsgrp.o + M = x S , y S if x = A y = A B A
4 1 hashprdifel S = 2 A S B S A B
5 simp1 A S B S A B A S
6 simp2 A S B S A B B S
7 5 5 6 3jca A S B S A B A S A S B S
8 4 7 syl S = 2 A S A S B S
9 simp3 A S B S A B A B
10 eqid + M = + M
11 1 2 3 10 mgm2nsgrplem2 A S B S A + M A + M B = A
12 11 3adant3 A S B S A B A + M A + M B = A
13 1 2 3 10 mgm2nsgrplem3 A S B S A + M A + M B = B
14 13 3adant3 A S B S A B A + M A + M B = B
15 9 12 14 3netr4d A S B S A B A + M A + M B A + M A + M B
16 4 15 syl S = 2 A + M A + M B A + M A + M B
17 2 eqcomi S = Base M
18 17 10 isnsgrp A S A S B S A + M A + M B A + M A + M B M Smgrp
19 8 16 18 sylc S = 2 M Smgrp