Metamath Proof Explorer


Theorem gsumvsmul

Description: Pull a scalar multiplication out of a sum of vectors. This theorem properly generalizes gsummulc2 , since every ring is a left module over itself. (Contributed by Stefan O'Rear, 6-Feb-2015) (Revised by Mario Carneiro, 5-May-2015) (Revised by AV, 10-Jul-2019)

Ref Expression
Hypotheses gsumvsmul.b B=BaseR
gsumvsmul.s S=ScalarR
gsumvsmul.k K=BaseS
gsumvsmul.z 0˙=0R
gsumvsmul.p +˙=+R
gsumvsmul.t ·˙=R
gsumvsmul.r φRLMod
gsumvsmul.a φAV
gsumvsmul.x φXK
gsumvsmul.y φkAYB
gsumvsmul.n φfinSupp0˙kAY
Assertion gsumvsmul φRkAX·˙Y=X·˙RkAY

Proof

Step Hyp Ref Expression
1 gsumvsmul.b B=BaseR
2 gsumvsmul.s S=ScalarR
3 gsumvsmul.k K=BaseS
4 gsumvsmul.z 0˙=0R
5 gsumvsmul.p +˙=+R
6 gsumvsmul.t ·˙=R
7 gsumvsmul.r φRLMod
8 gsumvsmul.a φAV
9 gsumvsmul.x φXK
10 gsumvsmul.y φkAYB
11 gsumvsmul.n φfinSupp0˙kAY
12 lmodcmn RLModRCMnd
13 7 12 syl φRCMnd
14 cmnmnd RCMndRMnd
15 13 14 syl φRMnd
16 1 2 6 3 lmodvsghm RLModXKyBX·˙yRGrpHomR
17 7 9 16 syl2anc φyBX·˙yRGrpHomR
18 ghmmhm yBX·˙yRGrpHomRyBX·˙yRMndHomR
19 17 18 syl φyBX·˙yRMndHomR
20 oveq2 y=YX·˙y=X·˙Y
21 oveq2 y=RkAYX·˙y=X·˙RkAY
22 1 4 13 15 8 19 10 11 20 21 gsummhm2 φRkAX·˙Y=X·˙RkAY