Metamath Proof Explorer


Theorem gsumvsmul1

Description: Pull a scalar multiplication out of a sum of vectors. This theorem properly generalizes gsummulc1 , since every ring is a left module over itself. (Contributed by Thierry Arnoux, 12-Jun-2023)

Ref Expression
Hypotheses gsumvsmul1.b B=BaseR
gsumvsmul1.s S=ScalarR
gsumvsmul1.k K=BaseS
gsumvsmul1.z 0˙=0S
gsumvsmul1.t ·˙=R
gsumvsmul1.r φRLMod
gsumvsmul1.1 φSCMnd
gsumvsmul1.a φAV
gsumvsmul1.x φYB
gsumvsmul1.y φkAXK
gsumvsmul1.n φfinSupp0˙kAX
Assertion gsumvsmul1 φRkAX·˙Y=SkAX·˙Y

Proof

Step Hyp Ref Expression
1 gsumvsmul1.b B=BaseR
2 gsumvsmul1.s S=ScalarR
3 gsumvsmul1.k K=BaseS
4 gsumvsmul1.z 0˙=0S
5 gsumvsmul1.t ·˙=R
6 gsumvsmul1.r φRLMod
7 gsumvsmul1.1 φSCMnd
8 gsumvsmul1.a φAV
9 gsumvsmul1.x φYB
10 gsumvsmul1.y φkAXK
11 gsumvsmul1.n φfinSupp0˙kAX
12 lmodcmn RLModRCMnd
13 cmnmnd RCMndRMnd
14 6 12 13 3syl φRMnd
15 1 2 5 3 lmodvslmhm RLModYBxKx·˙YSGrpHomR
16 6 9 15 syl2anc φxKx·˙YSGrpHomR
17 ghmmhm xKx·˙YSGrpHomRxKx·˙YSMndHomR
18 16 17 syl φxKx·˙YSMndHomR
19 oveq1 x=Xx·˙Y=X·˙Y
20 oveq1 x=SkAXx·˙Y=SkAX·˙Y
21 3 4 7 14 8 18 10 11 19 20 gsummhm2 φRkAX·˙Y=SkAX·˙Y