Metamath Proof Explorer


Theorem dib0

Description: The value of partial isomorphism B at the lattice zero is the singleton of the zero vector i.e. the zero subspace. (Contributed by NM, 27-Mar-2014)

Ref Expression
Hypotheses dib0.z 0 ˙ = 0. K
dib0.h H = LHyp K
dib0.i I = DIsoB K W
dib0.u U = DVecH K W
dib0.o O = 0 U
Assertion dib0 K HL W H I 0 ˙ = O

Proof

Step Hyp Ref Expression
1 dib0.z 0 ˙ = 0. K
2 dib0.h H = LHyp K
3 dib0.i I = DIsoB K W
4 dib0.u U = DVecH K W
5 dib0.o O = 0 U
6 fvex Base K V
7 resiexg Base K V I Base K V
8 6 7 ax-mp I Base K V
9 fvex LTrn K W V
10 9 mptex f LTrn K W I Base K V
11 8 10 xpsn I Base K × f LTrn K W I Base K = I Base K f LTrn K W I Base K
12 id K HL W H K HL W H
13 hlop K HL K OP
14 13 adantr K HL W H K OP
15 eqid Base K = Base K
16 15 1 op0cl K OP 0 ˙ Base K
17 14 16 syl K HL W H 0 ˙ Base K
18 15 2 lhpbase W H W Base K
19 eqid K = K
20 15 19 1 op0le K OP W Base K 0 ˙ K W
21 13 18 20 syl2an K HL W H 0 ˙ K W
22 eqid LTrn K W = LTrn K W
23 eqid f LTrn K W I Base K = f LTrn K W I Base K
24 eqid DIsoA K W = DIsoA K W
25 15 19 2 22 23 24 3 dibval2 K HL W H 0 ˙ Base K 0 ˙ K W I 0 ˙ = DIsoA K W 0 ˙ × f LTrn K W I Base K
26 12 17 21 25 syl12anc K HL W H I 0 ˙ = DIsoA K W 0 ˙ × f LTrn K W I Base K
27 15 1 2 24 dia0 K HL W H DIsoA K W 0 ˙ = I Base K
28 27 xpeq1d K HL W H DIsoA K W 0 ˙ × f LTrn K W I Base K = I Base K × f LTrn K W I Base K
29 26 28 eqtrd K HL W H I 0 ˙ = I Base K × f LTrn K W I Base K
30 15 2 22 4 5 23 dvh0g K HL W H O = I Base K f LTrn K W I Base K
31 30 sneqd K HL W H O = I Base K f LTrn K W I Base K
32 11 29 31 3eqtr4a K HL W H I 0 ˙ = O