Metamath Proof Explorer


Theorem mapdheq2biN

Description: Lemmma for ~? mapdh . Part (2) in Baer p. 45. The bidirectional version of mapdheq2 seems to require an additional hypothesis not mentioned in Baer. TODO fix ref. TODO: We probably don't need this; delete if never used. (Contributed by NM, 4-Apr-2015) (New usage is discouraged.)

Ref Expression
Hypotheses mapdh.q ⊢ Q = 0 C
mapdh.i ⊢ I = x ∈ V ⟼ if 2 nd ⁡ x = 0 ˙ Q ι h ∈ D | M ⁡ N ⁡ 2 nd ⁡ x = J ⁡ h ∧ M ⁡ N ⁡ 1 st ⁡ 1 st ⁡ x - ˙ 2 nd ⁡ x = J ⁡ 2 nd ⁡ 1 st ⁡ x R h
mapdh.h ⊢ H = LHyp ⁡ K
mapdh.m ⊢ M = mapd ⁡ K ⁡ W
mapdh.u ⊢ U = DVecH ⁡ K ⁡ W
mapdh.v ⊢ V = Base U
mapdh.s ⊢ - ˙ = - U
mapdhc.o ⊢ 0 ˙ = 0 U
mapdh.n ⊢ N = LSpan ⁡ U
mapdh.c ⊢ C = LCDual ⁡ K ⁡ W
mapdh.d ⊢ D = Base C
mapdh.r ⊢ R = - C
mapdh.j ⊢ J = LSpan ⁡ C
mapdh.k ⊢ φ → K ∈ HL ∧ W ∈ H
mapdhc.f ⊢ φ → F ∈ D
mapdh.mn ⊢ φ → M ⁡ N ⁡ X = J ⁡ F
mapdhcl.x ⊢ φ → X ∈ V ∖ 0 ˙
mapdhe2.y ⊢ φ → Y ∈ V ∖ 0 ˙
mapdhe2.g ⊢ φ → G ∈ D
mapdh.ne3 ⊢ φ → N ⁡ X ≠ N ⁡ Y
mapdh.my ⊢ φ → M ⁡ N ⁡ Y = J ⁡ G
Assertion mapdheq2biN ⊢ φ → I ⁡ X F Y = G ↔ I ⁡ Y G X = F

Proof

Step Hyp Ref Expression
1 mapdh.q ⊢ Q = 0 C
2 mapdh.i ⊢ I = x ∈ V ⟼ if 2 nd ⁡ x = 0 ˙ Q ι h ∈ D | M ⁡ N ⁡ 2 nd ⁡ x = J ⁡ h ∧ M ⁡ N ⁡ 1 st ⁡ 1 st ⁡ x - ˙ 2 nd ⁡ x = J ⁡ 2 nd ⁡ 1 st ⁡ x R h
3 mapdh.h ⊢ H = LHyp ⁡ K
4 mapdh.m ⊢ M = mapd ⁡ K ⁡ W
5 mapdh.u ⊢ U = DVecH ⁡ K ⁡ W
6 mapdh.v ⊢ V = Base U
7 mapdh.s ⊢ - ˙ = - U
8 mapdhc.o ⊢ 0 ˙ = 0 U
9 mapdh.n ⊢ N = LSpan ⁡ U
10 mapdh.c ⊢ C = LCDual ⁡ K ⁡ W
11 mapdh.d ⊢ D = Base C
12 mapdh.r ⊢ R = - C
13 mapdh.j ⊢ J = LSpan ⁡ C
14 mapdh.k ⊢ φ → K ∈ HL ∧ W ∈ H
15 mapdhc.f ⊢ φ → F ∈ D
16 mapdh.mn ⊢ φ → M ⁡ N ⁡ X = J ⁡ F
17 mapdhcl.x ⊢ φ → X ∈ V ∖ 0 ˙
18 mapdhe2.y ⊢ φ → Y ∈ V ∖ 0 ˙
19 mapdhe2.g ⊢ φ → G ∈ D
20 mapdh.ne3 ⊢ φ → N ⁡ X ≠ N ⁡ Y
21 mapdh.my ⊢ φ → M ⁡ N ⁡ Y = J ⁡ G
22 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 mapdheq2 ⊢ φ → I ⁡ X F Y = G → I ⁡ Y G X = F
23 20 necomd ⊢ φ → N ⁡ Y ≠ N ⁡ X
24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 19 21 18 17 15 23 mapdheq2 ⊢ φ → I ⁡ Y G X = F → I ⁡ X F Y = G
25 22 24 impbid ⊢ φ → I ⁡ X F Y = G ↔ I ⁡ Y G X = F