Metamath Proof Explorer


Theorem ismhm

Description: Property of a monoid homomorphism. (Contributed by Mario Carneiro, 7-Mar-2015)

Ref Expression
Hypotheses ismhm.b B=BaseS
ismhm.c C=BaseT
ismhm.p +˙=+S
ismhm.q ˙=+T
ismhm.z 0˙=0S
ismhm.y Y=0T
Assertion ismhm FSMndHomTSMndTMndF:BCxByBFx+˙y=Fx˙FyF0˙=Y

Proof

Step Hyp Ref Expression
1 ismhm.b B=BaseS
2 ismhm.c C=BaseT
3 ismhm.p +˙=+S
4 ismhm.q ˙=+T
5 ismhm.z 0˙=0S
6 ismhm.y Y=0T
7 df-mhm MndHom=sMnd,tMndfBasetBases|xBasesyBasesfx+sy=fx+tfyf0s=0t
8 7 elmpocl FSMndHomTSMndTMnd
9 fveq2 t=TBaset=BaseT
10 9 2 eqtr4di t=TBaset=C
11 fveq2 s=SBases=BaseS
12 11 1 eqtr4di s=SBases=B
13 10 12 oveqan12rd s=St=TBasetBases=CB
14 12 adantr s=St=TBases=B
15 fveq2 s=S+s=+S
16 15 3 eqtr4di s=S+s=+˙
17 16 oveqd s=Sx+sy=x+˙y
18 17 fveq2d s=Sfx+sy=fx+˙y
19 fveq2 t=T+t=+T
20 19 4 eqtr4di t=T+t=˙
21 20 oveqd t=Tfx+tfy=fx˙fy
22 18 21 eqeqan12d s=St=Tfx+sy=fx+tfyfx+˙y=fx˙fy
23 14 22 raleqbidv s=St=TyBasesfx+sy=fx+tfyyBfx+˙y=fx˙fy
24 14 23 raleqbidv s=St=TxBasesyBasesfx+sy=fx+tfyxByBfx+˙y=fx˙fy
25 fveq2 s=S0s=0S
26 25 5 eqtr4di s=S0s=0˙
27 26 fveq2d s=Sf0s=f0˙
28 fveq2 t=T0t=0T
29 28 6 eqtr4di t=T0t=Y
30 27 29 eqeqan12d s=St=Tf0s=0tf0˙=Y
31 24 30 anbi12d s=St=TxBasesyBasesfx+sy=fx+tfyf0s=0txByBfx+˙y=fx˙fyf0˙=Y
32 13 31 rabeqbidv s=St=TfBasetBases|xBasesyBasesfx+sy=fx+tfyf0s=0t=fCB|xByBfx+˙y=fx˙fyf0˙=Y
33 ovex CBV
34 33 rabex fCB|xByBfx+˙y=fx˙fyf0˙=YV
35 32 7 34 ovmpoa SMndTMndSMndHomT=fCB|xByBfx+˙y=fx˙fyf0˙=Y
36 35 eleq2d SMndTMndFSMndHomTFfCB|xByBfx+˙y=fx˙fyf0˙=Y
37 2 fvexi CV
38 1 fvexi BV
39 37 38 elmap FCBF:BC
40 39 anbi1i FCBxByBFx+˙y=Fx˙FyF0˙=YF:BCxByBFx+˙y=Fx˙FyF0˙=Y
41 fveq1 f=Ffx+˙y=Fx+˙y
42 fveq1 f=Ffx=Fx
43 fveq1 f=Ffy=Fy
44 42 43 oveq12d f=Ffx˙fy=Fx˙Fy
45 41 44 eqeq12d f=Ffx+˙y=fx˙fyFx+˙y=Fx˙Fy
46 45 2ralbidv f=FxByBfx+˙y=fx˙fyxByBFx+˙y=Fx˙Fy
47 fveq1 f=Ff0˙=F0˙
48 47 eqeq1d f=Ff0˙=YF0˙=Y
49 46 48 anbi12d f=FxByBfx+˙y=fx˙fyf0˙=YxByBFx+˙y=Fx˙FyF0˙=Y
50 49 elrab FfCB|xByBfx+˙y=fx˙fyf0˙=YFCBxByBFx+˙y=Fx˙FyF0˙=Y
51 3anass F:BCxByBFx+˙y=Fx˙FyF0˙=YF:BCxByBFx+˙y=Fx˙FyF0˙=Y
52 40 50 51 3bitr4i FfCB|xByBfx+˙y=fx˙fyf0˙=YF:BCxByBFx+˙y=Fx˙FyF0˙=Y
53 36 52 bitrdi SMndTMndFSMndHomTF:BCxByBFx+˙y=Fx˙FyF0˙=Y
54 8 53 biadanii FSMndHomTSMndTMndF:BCxByBFx+˙y=Fx˙FyF0˙=Y