Metamath Proof Explorer


Theorem grpidssd

Description: If the base set of a group is contained in the base set of another group, and the group operation of the group is the restriction of the group operation of the other group to its base set, then both groups have the same identity element. (Contributed by AV, 15-Mar-2019)

Ref Expression
Hypotheses grpidssd.m φMGrp
grpidssd.s φSGrp
grpidssd.b B=BaseS
grpidssd.c φBBaseM
grpidssd.o φxByBx+My=x+Sy
Assertion grpidssd φ0M=0S

Proof

Step Hyp Ref Expression
1 grpidssd.m φMGrp
2 grpidssd.s φSGrp
3 grpidssd.b B=BaseS
4 grpidssd.c φBBaseM
5 grpidssd.o φxByBx+My=x+Sy
6 eqid 0S=0S
7 3 6 grpidcl SGrp0SB
8 2 7 syl φ0SB
9 oveq1 x=0Sx+My=0S+My
10 oveq1 x=0Sx+Sy=0S+Sy
11 9 10 eqeq12d x=0Sx+My=x+Sy0S+My=0S+Sy
12 oveq2 y=0S0S+My=0S+M0S
13 oveq2 y=0S0S+Sy=0S+S0S
14 12 13 eqeq12d y=0S0S+My=0S+Sy0S+M0S=0S+S0S
15 11 14 rspc2va 0SB0SBxByBx+My=x+Sy0S+M0S=0S+S0S
16 8 8 5 15 syl21anc φ0S+M0S=0S+S0S
17 eqid +S=+S
18 3 17 6 grplid SGrp0SB0S+S0S=0S
19 2 7 18 syl2anc2 φ0S+S0S=0S
20 16 19 eqtrd φ0S+M0S=0S
21 4 8 sseldd φ0SBaseM
22 eqid BaseM=BaseM
23 eqid +M=+M
24 eqid 0M=0M
25 22 23 24 grpidlcan MGrp0SBaseM0SBaseM0S+M0S=0S0S=0M
26 1 21 21 25 syl3anc φ0S+M0S=0S0S=0M
27 20 26 mpbid φ0S=0M
28 27 eqcomd φ0M=0S