Metamath Proof Explorer


Theorem mapdpg

Description: Part 1 of proof of the first fundamental theorem of projective geometry. Part (1) in Baer p. 44. Our notation corresponds to Baer's as follows: M for *, N{ } for F(), J{ } for G(), X for x, G for x', Y for y, h for y'. TODO: Rename variables per mapdhval . (Contributed by NM, 22-Mar-2015)

Ref Expression
Hypotheses mapdpg.h H = LHyp K
mapdpg.m M = mapd K W
mapdpg.u U = DVecH K W
mapdpg.v V = Base U
mapdpg.s - ˙ = - U
mapdpg.z 0 ˙ = 0 U
mapdpg.n N = LSpan U
mapdpg.c C = LCDual K W
mapdpg.f F = Base C
mapdpg.r R = - C
mapdpg.j J = LSpan C
mapdpg.k φ K HL W H
mapdpg.x φ X V 0 ˙
mapdpg.y φ Y V 0 ˙
mapdpg.g φ G F
mapdpg.ne φ N X N Y
mapdpg.e φ M N X = J G
Assertion mapdpg φ ∃! h F M N Y = J h M N X - ˙ Y = J G R h

Proof

Step Hyp Ref Expression
1 mapdpg.h H = LHyp K
2 mapdpg.m M = mapd K W
3 mapdpg.u U = DVecH K W
4 mapdpg.v V = Base U
5 mapdpg.s - ˙ = - U
6 mapdpg.z 0 ˙ = 0 U
7 mapdpg.n N = LSpan U
8 mapdpg.c C = LCDual K W
9 mapdpg.f F = Base C
10 mapdpg.r R = - C
11 mapdpg.j J = LSpan C
12 mapdpg.k φ K HL W H
13 mapdpg.x φ X V 0 ˙
14 mapdpg.y φ Y V 0 ˙
15 mapdpg.g φ G F
16 mapdpg.ne φ N X N Y
17 mapdpg.e φ M N X = J G
18 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 mapdpglem24 φ h F M N Y = J h M N X - ˙ Y = J G R h
19 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 mapdpglem32 φ h F i F M N Y = J h M N X - ˙ Y = J G R h M N Y = J i M N X - ˙ Y = J G R i h = i
20 19 3exp φ h F i F M N Y = J h M N X - ˙ Y = J G R h M N Y = J i M N X - ˙ Y = J G R i h = i
21 20 ralrimivv φ h F i F M N Y = J h M N X - ˙ Y = J G R h M N Y = J i M N X - ˙ Y = J G R i h = i
22 sneq h = i h = i
23 22 fveq2d h = i J h = J i
24 23 eqeq2d h = i M N Y = J h M N Y = J i
25 oveq2 h = i G R h = G R i
26 25 sneqd h = i G R h = G R i
27 26 fveq2d h = i J G R h = J G R i
28 27 eqeq2d h = i M N X - ˙ Y = J G R h M N X - ˙ Y = J G R i
29 24 28 anbi12d h = i M N Y = J h M N X - ˙ Y = J G R h M N Y = J i M N X - ˙ Y = J G R i
30 29 reu4 ∃! h F M N Y = J h M N X - ˙ Y = J G R h h F M N Y = J h M N X - ˙ Y = J G R h h F i F M N Y = J h M N X - ˙ Y = J G R h M N Y = J i M N X - ˙ Y = J G R i h = i
31 18 21 30 sylanbrc φ ∃! h F M N Y = J h M N X - ˙ Y = J G R h