Metamath Proof Explorer


Theorem prjspreln0

Description: Two nonzero vectors are equivalent by a nonzero scalar. (Contributed by Steven Nguyen, 31-May-2023)

Ref Expression
Hypotheses prjsprel.1 ˙=xy|xByBlKx=l·˙y
prjspertr.b B=BaseV0V
prjspertr.s S=ScalarV
prjspertr.x ·˙=V
prjspertr.k K=BaseS
prjspreln0.z 0˙=0S
Assertion prjspreln0 VLVecX˙YXBYBmK0˙X=m·˙Y

Proof

Step Hyp Ref Expression
1 prjsprel.1 ˙=xy|xByBlKx=l·˙y
2 prjspertr.b B=BaseV0V
3 prjspertr.s S=ScalarV
4 prjspertr.x ·˙=V
5 prjspertr.k K=BaseS
6 prjspreln0.z 0˙=0S
7 1 prjsprel X˙YXBYBmKX=m·˙Y
8 simprl VLVecXBYBmKX=m·˙YmK
9 simplrl VLVecXBYBmKX=m·˙YXB
10 eldifsni XBaseV0VX0V
11 10 2 eleq2s XBX0V
12 9 11 syl VLVecXBYBmKX=m·˙YX0V
13 simplrr VLVecXBYBmKX=m·˙Ym=0˙X=m·˙Y
14 simpr VLVecXBYBmKX=m·˙Ym=0˙m=0˙
15 14 oveq1d VLVecXBYBmKX=m·˙Ym=0˙m·˙Y=0˙·˙Y
16 lveclmod VLVecVLMod
17 16 ad3antrrr VLVecXBYBmKX=m·˙Ym=0˙VLMod
18 difss BaseV0VBaseV
19 2 18 eqsstri BBaseV
20 simplrr VLVecXBYBmKX=m·˙Ym=0˙YB
21 20 anassrs VLVecXBYBmKX=m·˙Ym=0˙YB
22 19 21 sselid VLVecXBYBmKX=m·˙Ym=0˙YBaseV
23 eqid BaseV=BaseV
24 eqid 0V=0V
25 23 3 4 6 24 lmod0vs VLModYBaseV0˙·˙Y=0V
26 17 22 25 syl2anc VLVecXBYBmKX=m·˙Ym=0˙0˙·˙Y=0V
27 13 15 26 3eqtrd VLVecXBYBmKX=m·˙Ym=0˙X=0V
28 12 27 mteqand VLVecXBYBmKX=m·˙Ym0˙
29 nelsn m0˙¬m0˙
30 28 29 syl VLVecXBYBmKX=m·˙Y¬m0˙
31 8 30 eldifd VLVecXBYBmKX=m·˙YmK0˙
32 31 ex VLVecXBYBmKX=m·˙YmK0˙
33 simpr mKX=m·˙YX=m·˙Y
34 32 33 jca2 VLVecXBYBmKX=m·˙YmK0˙X=m·˙Y
35 34 reximdv2 VLVecXBYBmKX=m·˙YmK0˙X=m·˙Y
36 difss K0˙K
37 ssrexv K0˙KmK0˙X=m·˙YmKX=m·˙Y
38 36 37 mp1i VLVecXBYBmK0˙X=m·˙YmKX=m·˙Y
39 35 38 impbid VLVecXBYBmKX=m·˙YmK0˙X=m·˙Y
40 39 pm5.32da VLVecXBYBmKX=m·˙YXBYBmK0˙X=m·˙Y
41 7 40 bitrid VLVecX˙YXBYBmK0˙X=m·˙Y