Metamath Proof Explorer


Theorem prjsperref

Description: The relation in PrjSp is reflexive. (Contributed by Steven Nguyen, 30-Apr-2023)

Ref Expression
Hypotheses prjsprel.1 ˙ = x y | x B y B l K x = l · ˙ y
prjspertr.b B = Base V 0 V
prjspertr.s S = Scalar V
prjspertr.x · ˙ = V
prjspertr.k K = Base S
Assertion prjsperref V LMod X B X ˙ X

Proof

Step Hyp Ref Expression
1 prjsprel.1 ˙ = x y | x B y B l K x = l · ˙ y
2 prjspertr.b B = Base V 0 V
3 prjspertr.s S = Scalar V
4 prjspertr.x · ˙ = V
5 prjspertr.k K = Base S
6 eqid 1 S = 1 S
7 3 5 6 lmod1cl V LMod 1 S K
8 7 adantr V LMod X B 1 S K
9 oveq1 m = 1 S m · ˙ X = 1 S · ˙ X
10 9 eqeq2d m = 1 S X = m · ˙ X X = 1 S · ˙ X
11 10 adantl V LMod X B m = 1 S X = m · ˙ X X = 1 S · ˙ X
12 eldifi X Base V 0 V X Base V
13 12 2 eleq2s X B X Base V
14 eqid Base V = Base V
15 14 3 4 6 lmodvs1 V LMod X Base V 1 S · ˙ X = X
16 13 15 sylan2 V LMod X B 1 S · ˙ X = X
17 16 eqcomd V LMod X B X = 1 S · ˙ X
18 8 11 17 rspcedvd V LMod X B m K X = m · ˙ X
19 18 ex V LMod X B m K X = m · ˙ X
20 19 pm4.71d V LMod X B X B m K X = m · ˙ X
21 pm4.24 X B X B X B
22 21 anbi1i X B m K X = m · ˙ X X B X B m K X = m · ˙ X
23 1 prjsprel X ˙ X X B X B m K X = m · ˙ X
24 22 23 bitr4i X B m K X = m · ˙ X X ˙ X
25 20 24 bitrdi V LMod X B X ˙ X