Metamath Proof Explorer


Theorem dibelval3

Description: Member of the partial isomorphism B. (Contributed by NM, 26-Feb-2014)

Ref Expression
Hypotheses dibval3.b B=BaseK
dibval3.l ˙=K
dibval3.h H=LHypK
dibval3.t T=LTrnKW
dibval3.r R=trLKW
dibval3.o 0˙=gTIB
dibval3.i I=DIsoBKW
Assertion dibelval3 KVWHXBX˙WYIXfTY=f0˙Rf˙X

Proof

Step Hyp Ref Expression
1 dibval3.b B=BaseK
2 dibval3.l ˙=K
3 dibval3.h H=LHypK
4 dibval3.t T=LTrnKW
5 dibval3.r R=trLKW
6 dibval3.o 0˙=gTIB
7 dibval3.i I=DIsoBKW
8 eqid DIsoAKW=DIsoAKW
9 1 2 3 4 6 8 7 dibval2 KVWHXBX˙WIX=DIsoAKWX×0˙
10 9 eleq2d KVWHXBX˙WYIXYDIsoAKWX×0˙
11 1 2 3 4 5 8 diaelval KVWHXBX˙WfDIsoAKWXfTRf˙X
12 11 anbi1d KVWHXBX˙WfDIsoAKWXY=f0˙fTRf˙XY=f0˙
13 an13 Y=fsfDIsoAKWXs0˙s0˙fDIsoAKWXY=fs
14 velsn s0˙s=0˙
15 14 anbi1i s0˙fDIsoAKWXY=fss=0˙fDIsoAKWXY=fs
16 13 15 bitri Y=fsfDIsoAKWXs0˙s=0˙fDIsoAKWXY=fs
17 16 exbii sY=fsfDIsoAKWXs0˙ss=0˙fDIsoAKWXY=fs
18 4 fvexi TV
19 18 mptex gTIBV
20 6 19 eqeltri 0˙V
21 opeq2 s=0˙fs=f0˙
22 21 eqeq2d s=0˙Y=fsY=f0˙
23 22 anbi2d s=0˙fDIsoAKWXY=fsfDIsoAKWXY=f0˙
24 20 23 ceqsexv ss=0˙fDIsoAKWXY=fsfDIsoAKWXY=f0˙
25 17 24 bitri sY=fsfDIsoAKWXs0˙fDIsoAKWXY=f0˙
26 anass fTY=f0˙Rf˙XfTY=f0˙Rf˙X
27 an32 fTY=f0˙Rf˙XfTRf˙XY=f0˙
28 26 27 bitr3i fTY=f0˙Rf˙XfTRf˙XY=f0˙
29 12 25 28 3bitr4g KVWHXBX˙WsY=fsfDIsoAKWXs0˙fTY=f0˙Rf˙X
30 29 exbidv KVWHXBX˙WfsY=fsfDIsoAKWXs0˙ffTY=f0˙Rf˙X
31 elxp YDIsoAKWX×0˙fsY=fsfDIsoAKWXs0˙
32 df-rex fTY=f0˙Rf˙XffTY=f0˙Rf˙X
33 30 31 32 3bitr4g KVWHXBX˙WYDIsoAKWX×0˙fTY=f0˙Rf˙X
34 10 33 bitrd KVWHXBX˙WYIXfTY=f0˙Rf˙X