Metamath Proof Explorer


Theorem dihlspsnssN

Description: A subspace included in a 1-dim subspace belongs to the range of isomorphism H. (Contributed by NM, 26-Apr-2014) (New usage is discouraged.)

Ref Expression
Hypotheses dih1dor0.h ⊢ H = LHyp ⁡ K
dih1dor0.u ⊢ U = DVecH ⁡ K ⁡ W
dihldor0.v ⊢ V = Base U
dih1dor0.s ⊢ S = LSubSp ⁡ U
dih1dor0.n ⊢ N = LSpan ⁡ U
dih1dor0.i ⊢ I = DIsoH ⁡ K ⁡ W
Assertion dihlspsnssN ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X → T ∈ S ↔ T ∈ ran ⁡ I

Proof

Step Hyp Ref Expression
1 dih1dor0.h ⊢ H = LHyp ⁡ K
2 dih1dor0.u ⊢ U = DVecH ⁡ K ⁡ W
3 dihldor0.v ⊢ V = Base U
4 dih1dor0.s ⊢ S = LSubSp ⁡ U
5 dih1dor0.n ⊢ N = LSpan ⁡ U
6 dih1dor0.i ⊢ I = DIsoH ⁡ K ⁡ W
7 simpr ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = N ⁡ X → T = N ⁡ X
8 1 2 3 5 6 dihlsprn ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V → N ⁡ X ∈ ran ⁡ I
9 8 3adant3 ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X → N ⁡ X ∈ ran ⁡ I
10 9 ad2antrr ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = N ⁡ X → N ⁡ X ∈ ran ⁡ I
11 7 10 eqeltrd ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = N ⁡ X → T ∈ ran ⁡ I
12 simpr ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → T = 0 U
13 simpll1 ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → K ∈ HL ∧ W ∈ H
14 eqid ⊢ 0. ⁡ K = 0. ⁡ K
15 eqid ⊢ 0 U = 0 U
16 14 1 6 2 15 dih0 ⊢ K ∈ HL ∧ W ∈ H → I ⁡ 0. ⁡ K = 0 U
17 13 16 syl ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → I ⁡ 0. ⁡ K = 0 U
18 12 17 eqtr4d ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → T = I ⁡ 0. ⁡ K
19 eqid ⊢ Base K = Base K
20 19 1 6 dihfn ⊢ K ∈ HL ∧ W ∈ H → I Fn Base K
21 13 20 syl ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → I Fn Base K
22 simp1l ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X → K ∈ HL
23 22 ad2antrr ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → K ∈ HL
24 hlop ⊢ K ∈ HL → K ∈ OP
25 19 14 op0cl ⊢ K ∈ OP → 0. ⁡ K ∈ Base K
26 23 24 25 3syl ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → 0. ⁡ K ∈ Base K
27 fnfvelrn ⊢ I Fn Base K ∧ 0. ⁡ K ∈ Base K → I ⁡ 0. ⁡ K ∈ ran ⁡ I
28 21 26 27 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → I ⁡ 0. ⁡ K ∈ ran ⁡ I
29 18 28 eqeltrd ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S ∧ T = 0 U → T ∈ ran ⁡ I
30 simpl1 ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S → K ∈ HL ∧ W ∈ H
31 1 2 30 dvhlvec ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S → U ∈ LVec
32 simpr ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S → T ∈ S
33 simpl2 ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S → X ∈ V
34 simpl3 ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S → T ⊆ N ⁡ X
35 3 15 4 5 lspsnat ⊢ U ∈ LVec ∧ T ∈ S ∧ X ∈ V ∧ T ⊆ N ⁡ X → T = N ⁡ X ∨ T = 0 U
36 31 32 33 34 35 syl31anc ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S → T = N ⁡ X ∨ T = 0 U
37 11 29 36 mpjaodan ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X ∧ T ∈ S → T ∈ ran ⁡ I
38 37 ex ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X → T ∈ S → T ∈ ran ⁡ I
39 1 2 6 4 dihsslss ⊢ K ∈ HL ∧ W ∈ H → ran ⁡ I ⊆ S
40 39 3ad2ant1 ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X → ran ⁡ I ⊆ S
41 40 sseld ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X → T ∈ ran ⁡ I → T ∈ S
42 38 41 impbid ⊢ K ∈ HL ∧ W ∈ H ∧ X ∈ V ∧ T ⊆ N ⁡ X → T ∈ S ↔ T ∈ ran ⁡ I