Metamath Proof Explorer


Theorem submnd0

Description: The zero of a submonoid is the same as the zero in the parent monoid. (Note that we must add the condition that the zero of the parent monoid is actually contained in the submonoid, because it is possible to have "subsets that are monoids" which are not submonoids because they have a different identity element. See, for example, smndex1mnd and smndex1n0mnd ). (Contributed by Mario Carneiro, 10-Jan-2015)

Ref Expression
Hypotheses submnd0.b B=BaseG
submnd0.z 0˙=0G
submnd0.h H=G𝑠S
Assertion submnd0 GMndHMndSB0˙S0˙=0H

Proof

Step Hyp Ref Expression
1 submnd0.b B=BaseG
2 submnd0.z 0˙=0G
3 submnd0.h H=G𝑠S
4 eqid BaseH=BaseH
5 eqid 0H=0H
6 eqid +H=+H
7 simprr GMndHMndSB0˙S0˙S
8 3 1 ressbas2 SBS=BaseH
9 8 ad2antrl GMndHMndSB0˙SS=BaseH
10 7 9 eleqtrd GMndHMndSB0˙S0˙BaseH
11 fvex BaseHV
12 9 11 eqeltrdi GMndHMndSB0˙SSV
13 12 adantr GMndHMndSB0˙SxBaseHSV
14 eqid +G=+G
15 3 14 ressplusg SV+G=+H
16 13 15 syl GMndHMndSB0˙SxBaseH+G=+H
17 16 oveqd GMndHMndSB0˙SxBaseH0˙+Gx=0˙+Hx
18 simpll GMndHMndSB0˙SGMnd
19 3 1 ressbasss BaseHB
20 19 sseli xBaseHxB
21 1 14 2 mndlid GMndxB0˙+Gx=x
22 18 20 21 syl2an GMndHMndSB0˙SxBaseH0˙+Gx=x
23 17 22 eqtr3d GMndHMndSB0˙SxBaseH0˙+Hx=x
24 16 oveqd GMndHMndSB0˙SxBaseHx+G0˙=x+H0˙
25 1 14 2 mndrid GMndxBx+G0˙=x
26 18 20 25 syl2an GMndHMndSB0˙SxBaseHx+G0˙=x
27 24 26 eqtr3d GMndHMndSB0˙SxBaseHx+H0˙=x
28 4 5 6 10 23 27 ismgmid2 GMndHMndSB0˙S0˙=0H