Metamath Proof Explorer


Theorem hdmap14lem7

Description: Combine cases of F . TODO: Can this be done at once in hdmap14lem3 , in order to get rid of hdmap14lem6 ? Perhaps modify lspsneu to become E! k e. K instead of E! k e. ( K \ { .0. } ) ? (Contributed by NM, 1-Jun-2015)

Ref Expression
Hypotheses hdmap14lem7.h H = LHyp K
hdmap14lem7.u U = DVecH K W
hdmap14lem7.v V = Base U
hdmap14lem7.t · ˙ = U
hdmap14lem7.o 0 ˙ = 0 U
hdmap14lem7.r R = Scalar U
hdmap14lem7.b B = Base R
hdmap14lem7.c C = LCDual K W
hdmap14lem7.e ˙ = C
hdmap14lem7.p P = Scalar C
hdmap14lem7.a A = Base P
hdmap14lem7.s S = HDMap K W
hdmap14lem7.k φ K HL W H
hdmap14lem7.x φ X V 0 ˙
hdmap14lem7.f φ F B
Assertion hdmap14lem7 φ ∃! g A S F · ˙ X = g ˙ S X

Proof

Step Hyp Ref Expression
1 hdmap14lem7.h H = LHyp K
2 hdmap14lem7.u U = DVecH K W
3 hdmap14lem7.v V = Base U
4 hdmap14lem7.t · ˙ = U
5 hdmap14lem7.o 0 ˙ = 0 U
6 hdmap14lem7.r R = Scalar U
7 hdmap14lem7.b B = Base R
8 hdmap14lem7.c C = LCDual K W
9 hdmap14lem7.e ˙ = C
10 hdmap14lem7.p P = Scalar C
11 hdmap14lem7.a A = Base P
12 hdmap14lem7.s S = HDMap K W
13 hdmap14lem7.k φ K HL W H
14 hdmap14lem7.x φ X V 0 ˙
15 hdmap14lem7.f φ F B
16 eqid 0 R = 0 R
17 eqid LSpan C = LSpan C
18 eqid 0 P = 0 P
19 13 adantr φ F = 0 R K HL W H
20 14 adantr φ F = 0 R X V 0 ˙
21 simpr φ F = 0 R F = 0 R
22 1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 19 20 21 hdmap14lem6 φ F = 0 R ∃! g A S F · ˙ X = g ˙ S X
23 13 adantr φ F 0 R K HL W H
24 14 adantr φ F 0 R X V 0 ˙
25 15 adantr φ F 0 R F B
26 simpr φ F 0 R F 0 R
27 eldifsn F B 0 R F B F 0 R
28 25 26 27 sylanbrc φ F 0 R F B 0 R
29 1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 23 24 28 hdmap14lem4 φ F 0 R ∃! g A S F · ˙ X = g ˙ S X
30 22 29 pm2.61dane φ ∃! g A S F · ˙ X = g ˙ S X