Metamath Proof Explorer


Theorem lcosslsp

Description: Lemma for lspeqlco . (Contributed by AV, 20-Apr-2019)

Ref Expression
Hypothesis lspeqvlco.b B = Base M
Assertion lcosslsp M LMod V 𝒫 B M LinCo V LSpan M V

Proof

Step Hyp Ref Expression
1 lspeqvlco.b B = Base M
2 ellcoellss M LMod s LSubSp M V s y M LinCo V y s
3 2 3exp M LMod s LSubSp M V s y M LinCo V y s
4 3 ad2antrr M LMod V 𝒫 B x M LinCo V s LSubSp M V s y M LinCo V y s
5 4 imp M LMod V 𝒫 B x M LinCo V s LSubSp M V s y M LinCo V y s
6 elequ1 y = x y s x s
7 6 rspcv x M LinCo V y M LinCo V y s x s
8 7 ad2antlr M LMod V 𝒫 B x M LinCo V s LSubSp M y M LinCo V y s x s
9 5 8 syld M LMod V 𝒫 B x M LinCo V s LSubSp M V s x s
10 9 ralrimiva M LMod V 𝒫 B x M LinCo V s LSubSp M V s x s
11 vex x V
12 11 elintrab x s LSubSp M | V s s LSubSp M V s x s
13 10 12 sylibr M LMod V 𝒫 B x M LinCo V x s LSubSp M | V s
14 simpll M LMod V 𝒫 B x M LinCo V M LMod
15 elpwi V 𝒫 B V B
16 15 ad2antlr M LMod V 𝒫 B x M LinCo V V B
17 eqid LSubSp M = LSubSp M
18 eqid LSpan M = LSpan M
19 1 17 18 lspval M LMod V B LSpan M V = s LSubSp M | V s
20 14 16 19 syl2anc M LMod V 𝒫 B x M LinCo V LSpan M V = s LSubSp M | V s
21 13 20 eleqtrrd M LMod V 𝒫 B x M LinCo V x LSpan M V
22 21 ex M LMod V 𝒫 B x M LinCo V x LSpan M V
23 22 ssrdv M LMod V 𝒫 B M LinCo V LSpan M V