Metamath Proof Explorer


Theorem hdmap1cl

Description: Convert closure theorem mapdhcl to use HDMap1 function. (Contributed by NM, 15-May-2015)

Ref Expression
Hypotheses hdmap1eq2.h H = LHyp K
hdmap1eq2.u U = DVecH K W
hdmap1eq2.v V = Base U
hdmap1eq2.o 0 ˙ = 0 U
hdmap1eq2.n N = LSpan U
hdmap1eq2.c C = LCDual K W
hdmap1eq2.d D = Base C
hdmap1eq2.l L = LSpan C
hdmap1eq2.m M = mapd K W
hdmap1eq2.i I = HDMap1 K W
hdmap1eq2.k φ K HL W H
hdmap1eq2.f φ F D
hdmap1eq2.mn φ M N X = L F
hdmap1cl.ne φ N X N Y
hdmap1cl.x φ X V 0 ˙
hdmap1cl.y φ Y V
Assertion hdmap1cl φ I X F Y D

Proof

Step Hyp Ref Expression
1 hdmap1eq2.h H = LHyp K
2 hdmap1eq2.u U = DVecH K W
3 hdmap1eq2.v V = Base U
4 hdmap1eq2.o 0 ˙ = 0 U
5 hdmap1eq2.n N = LSpan U
6 hdmap1eq2.c C = LCDual K W
7 hdmap1eq2.d D = Base C
8 hdmap1eq2.l L = LSpan C
9 hdmap1eq2.m M = mapd K W
10 hdmap1eq2.i I = HDMap1 K W
11 hdmap1eq2.k φ K HL W H
12 hdmap1eq2.f φ F D
13 hdmap1eq2.mn φ M N X = L F
14 hdmap1cl.ne φ N X N Y
15 hdmap1cl.x φ X V 0 ˙
16 hdmap1cl.y φ Y V
17 eqid - U = - U
18 eqid - C = - C
19 eqid 0 C = 0 C
20 eqid x V if 2 nd x = 0 ˙ 0 C ι h D | M N 2 nd x = L h M N 1 st 1 st x - U 2 nd x = L 2 nd 1 st x - C h = x V if 2 nd x = 0 ˙ 0 C ι h D | M N 2 nd x = L h M N 1 st 1 st x - U 2 nd x = L 2 nd 1 st x - C h
21 1 2 3 17 4 5 6 7 18 19 8 9 10 11 15 12 16 20 hdmap1valc φ I X F Y = x V if 2 nd x = 0 ˙ 0 C ι h D | M N 2 nd x = L h M N 1 st 1 st x - U 2 nd x = L 2 nd 1 st x - C h X F Y
22 19 20 1 9 2 3 17 4 5 6 7 18 8 11 12 13 15 16 14 mapdhcl φ x V if 2 nd x = 0 ˙ 0 C ι h D | M N 2 nd x = L h M N 1 st 1 st x - U 2 nd x = L 2 nd 1 st x - C h X F Y D
23 21 22 eqeltrd φ I X F Y D