Metamath Proof Explorer


Theorem hdmap1eq2

Description: Convert mapdheq2 to use HDMap1 function. (Contributed by NM, 16-May-2015)

Ref Expression
Hypotheses hdmap1eq2.h H=LHypK
hdmap1eq2.u U=DVecHKW
hdmap1eq2.v V=BaseU
hdmap1eq2.o 0˙=0U
hdmap1eq2.n N=LSpanU
hdmap1eq2.c C=LCDualKW
hdmap1eq2.d D=BaseC
hdmap1eq2.l L=LSpanC
hdmap1eq2.m M=mapdKW
hdmap1eq2.i I=HDMap1KW
hdmap1eq2.k φKHLWH
hdmap1eq2.f φFD
hdmap1eq2.mn φMNX=LF
hdmap1eq2.ne φNXNY
hdmap1eq2.x φXV0˙
hdmap1eq2.y φYV0˙
hdmap1eq2.e φIXFY=G
Assertion hdmap1eq2 φIYGX=F

Proof

Step Hyp Ref Expression
1 hdmap1eq2.h H=LHypK
2 hdmap1eq2.u U=DVecHKW
3 hdmap1eq2.v V=BaseU
4 hdmap1eq2.o 0˙=0U
5 hdmap1eq2.n N=LSpanU
6 hdmap1eq2.c C=LCDualKW
7 hdmap1eq2.d D=BaseC
8 hdmap1eq2.l L=LSpanC
9 hdmap1eq2.m M=mapdKW
10 hdmap1eq2.i I=HDMap1KW
11 hdmap1eq2.k φKHLWH
12 hdmap1eq2.f φFD
13 hdmap1eq2.mn φMNX=LF
14 hdmap1eq2.ne φNXNY
15 hdmap1eq2.x φXV0˙
16 hdmap1eq2.y φYV0˙
17 hdmap1eq2.e φIXFY=G
18 eqid -U=-U
19 eqid -C=-C
20 eqid 0C=0C
21 16 eldifad φYV
22 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 21 hdmap1cl φIXFYD
23 17 22 eqeltrrd φGD
24 15 eldifad φXV
25 eqid xVif2ndx=0˙0CιhD|MN2ndx=LhMN1st1stx-U2ndx=L2nd1stx-Ch=xVif2ndx=0˙0CιhD|MN2ndx=LhMN1st1stx-U2ndx=L2nd1stx-Ch
26 1 2 3 18 4 5 6 7 19 20 8 9 10 11 16 23 24 25 hdmap1valc φIYGX=xVif2ndx=0˙0CιhD|MN2ndx=LhMN1st1stx-U2ndx=L2nd1stx-ChYGX
27 1 2 3 18 4 5 6 7 19 20 8 9 10 11 15 12 21 25 hdmap1valc φIXFY=xVif2ndx=0˙0CιhD|MN2ndx=LhMN1st1stx-U2ndx=L2nd1stx-ChXFY
28 27 17 eqtr3d φxVif2ndx=0˙0CιhD|MN2ndx=LhMN1st1stx-U2ndx=L2nd1stx-ChXFY=G
29 1 2 3 18 4 5 6 7 19 20 8 9 25 11 12 13 28 14 15 16 mapdh75e φxVif2ndx=0˙0CιhD|MN2ndx=LhMN1st1stx-U2ndx=L2nd1stx-ChYGX=F
30 26 29 eqtrd φIYGX=F