Metamath Proof Explorer


Theorem prjspnerlem

Description: A lemma showing that the equivalence relation used in prjspnval2 and the equivalence relation used in prjspval are equal, but only with the antecedent K e. DivRing . (Contributed by SN, 15-Jul-2023)

Ref Expression
Hypotheses prjspnerlem.e ˙ = x y | x B y B l S x = l · ˙ y
prjspnerlem.w W = K freeLMod 0 N
prjspnerlem.b B = Base W 0 W
prjspnerlem.s S = Base K
prjspnerlem.x · ˙ = W
Assertion prjspnerlem K DivRing ˙ = x y | x B y B l Base Scalar W x = l · ˙ y

Proof

Step Hyp Ref Expression
1 prjspnerlem.e ˙ = x y | x B y B l S x = l · ˙ y
2 prjspnerlem.w W = K freeLMod 0 N
3 prjspnerlem.b B = Base W 0 W
4 prjspnerlem.s S = Base K
5 prjspnerlem.x · ˙ = W
6 ovex 0 N V
7 2 frlmsca K DivRing 0 N V K = Scalar W
8 6 7 mpan2 K DivRing K = Scalar W
9 8 fveq2d K DivRing Base K = Base Scalar W
10 4 9 syl5eq K DivRing S = Base Scalar W
11 10 rexeqdv K DivRing l S x = l · ˙ y l Base Scalar W x = l · ˙ y
12 11 anbi2d K DivRing x B y B l S x = l · ˙ y x B y B l Base Scalar W x = l · ˙ y
13 12 opabbidv K DivRing x y | x B y B l S x = l · ˙ y = x y | x B y B l Base Scalar W x = l · ˙ y
14 1 13 syl5eq K DivRing ˙ = x y | x B y B l Base Scalar W x = l · ˙ y