Metamath Proof Explorer


Theorem hdmapevec2

Description: The inner product of the reference vector E with itself is nonzero. This shows the inner product condition in the proof of Theorem 3.6 of Holland95 p. 14 line 32, [ e , e ] =/= 0 is satisfied. TODO: remove redundant hypothesis hdmapevec.j. (Contributed by NM, 1-Jun-2015)

Ref Expression
Hypotheses hdmapevec.h H = LHyp K
hdmapevec.e E = I Base K I LTrn K W
hdmapevec.j J = HVMap K W
hdmapevec.s S = HDMap K W
hdmapevec.k φ K HL W H
hdmapevec2.u U = DVecH K W
hdmapevec2.r R = Scalar U
hdmapevec2.i 1 ˙ = 1 R
Assertion hdmapevec2 φ S E E = 1 ˙

Proof

Step Hyp Ref Expression
1 hdmapevec.h H = LHyp K
2 hdmapevec.e E = I Base K I LTrn K W
3 hdmapevec.j J = HVMap K W
4 hdmapevec.s S = HDMap K W
5 hdmapevec.k φ K HL W H
6 hdmapevec2.u U = DVecH K W
7 hdmapevec2.r R = Scalar U
8 hdmapevec2.i 1 ˙ = 1 R
9 1 2 3 4 5 hdmapevec φ S E = J E
10 eqid ocH K W = ocH K W
11 eqid Base U = Base U
12 eqid + U = + U
13 eqid U = U
14 eqid 0 U = 0 U
15 eqid Base R = Base R
16 eqid Base K = Base K
17 eqid LTrn K W = LTrn K W
18 1 16 17 6 11 14 2 5 dvheveccl φ E Base U 0 U
19 1 6 10 11 12 13 14 7 15 3 5 18 hvmapval φ J E = v Base U ι k Base R | w ocH K W E v = w + U k U E
20 9 19 eqtrd φ S E = v Base U ι k Base R | w ocH K W E v = w + U k U E
21 20 fveq1d φ S E E = v Base U ι k Base R | w ocH K W E v = w + U k U E E
22 eqid v Base U ι k Base R | w ocH K W E v = w + U k U E = v Base U ι k Base R | w ocH K W E v = w + U k U E
23 1 10 6 11 12 13 14 7 15 8 5 18 22 dochfl1 φ v Base U ι k Base R | w ocH K W E v = w + U k U E E = 1 ˙
24 21 23 eqtrd φ S E E = 1 ˙