Metamath Proof Explorer


Theorem linindslinci

Description: The implications of being a linearly independent subset and a linear combination of this subset being 0. (Contributed by AV, 24-Apr-2019) (Revised by AV, 30-Jul-2019)

Ref Expression
Hypotheses islininds.b B = Base M
islininds.z Z = 0 M
islininds.r R = Scalar M
islininds.e E = Base R
islininds.0 0 ˙ = 0 R
Assertion linindslinci S linIndS M F E S finSupp 0 ˙ F F linC M S = Z x S F x = 0 ˙

Proof

Step Hyp Ref Expression
1 islininds.b B = Base M
2 islininds.z Z = 0 M
3 islininds.r R = Scalar M
4 islininds.e E = Base R
5 islininds.0 0 ˙ = 0 R
6 1 2 3 4 5 linindsi S linIndS M S 𝒫 B f E S finSupp 0 ˙ f f linC M S = Z x S f x = 0 ˙
7 breq1 f = F finSupp 0 ˙ f finSupp 0 ˙ F
8 oveq1 f = F f linC M S = F linC M S
9 8 eqeq1d f = F f linC M S = Z F linC M S = Z
10 7 9 anbi12d f = F finSupp 0 ˙ f f linC M S = Z finSupp 0 ˙ F F linC M S = Z
11 fveq1 f = F f x = F x
12 11 eqeq1d f = F f x = 0 ˙ F x = 0 ˙
13 12 ralbidv f = F x S f x = 0 ˙ x S F x = 0 ˙
14 10 13 imbi12d f = F finSupp 0 ˙ f f linC M S = Z x S f x = 0 ˙ finSupp 0 ˙ F F linC M S = Z x S F x = 0 ˙
15 14 rspcv F E S f E S finSupp 0 ˙ f f linC M S = Z x S f x = 0 ˙ finSupp 0 ˙ F F linC M S = Z x S F x = 0 ˙
16 15 com23 F E S finSupp 0 ˙ F F linC M S = Z f E S finSupp 0 ˙ f f linC M S = Z x S f x = 0 ˙ x S F x = 0 ˙
17 16 3impib F E S finSupp 0 ˙ F F linC M S = Z f E S finSupp 0 ˙ f f linC M S = Z x S f x = 0 ˙ x S F x = 0 ˙
18 17 com12 f E S finSupp 0 ˙ f f linC M S = Z x S f x = 0 ˙ F E S finSupp 0 ˙ F F linC M S = Z x S F x = 0 ˙
19 6 18 simpl2im S linIndS M F E S finSupp 0 ˙ F F linC M S = Z x S F x = 0 ˙
20 19 imp S linIndS M F E S finSupp 0 ˙ F F linC M S = Z x S F x = 0 ˙