Metamath Proof Explorer


Theorem mat2pmatfval

Description: Value of the matrix transformation. (Contributed by AV, 31-Jul-2019)

Ref Expression
Hypotheses mat2pmatfval.t T = N matToPolyMat R
mat2pmatfval.a A = N Mat R
mat2pmatfval.b B = Base A
mat2pmatfval.p P = Poly 1 R
mat2pmatfval.s S = algSc P
Assertion mat2pmatfval N Fin R V T = m B x N , y N S x m y

Proof

Step Hyp Ref Expression
1 mat2pmatfval.t T = N matToPolyMat R
2 mat2pmatfval.a A = N Mat R
3 mat2pmatfval.b B = Base A
4 mat2pmatfval.p P = Poly 1 R
5 mat2pmatfval.s S = algSc P
6 df-mat2pmat matToPolyMat = n Fin , r V m Base n Mat r x n , y n algSc Poly 1 r x m y
7 6 a1i N Fin R V matToPolyMat = n Fin , r V m Base n Mat r x n , y n algSc Poly 1 r x m y
8 oveq12 n = N r = R n Mat r = N Mat R
9 8 fveq2d n = N r = R Base n Mat r = Base N Mat R
10 2 fveq2i Base A = Base N Mat R
11 3 10 eqtr2i Base N Mat R = B
12 9 11 eqtrdi n = N r = R Base n Mat r = B
13 simpl n = N r = R n = N
14 2fveq3 r = R algSc Poly 1 r = algSc Poly 1 R
15 14 adantl n = N r = R algSc Poly 1 r = algSc Poly 1 R
16 4 fveq2i algSc P = algSc Poly 1 R
17 5 16 eqtr2i algSc Poly 1 R = S
18 15 17 eqtrdi n = N r = R algSc Poly 1 r = S
19 18 fveq1d n = N r = R algSc Poly 1 r x m y = S x m y
20 13 13 19 mpoeq123dv n = N r = R x n , y n algSc Poly 1 r x m y = x N , y N S x m y
21 12 20 mpteq12dv n = N r = R m Base n Mat r x n , y n algSc Poly 1 r x m y = m B x N , y N S x m y
22 21 adantl N Fin R V n = N r = R m Base n Mat r x n , y n algSc Poly 1 r x m y = m B x N , y N S x m y
23 simpl N Fin R V N Fin
24 elex R V R V
25 24 adantl N Fin R V R V
26 3 fvexi B V
27 mptexg B V m B x N , y N S x m y V
28 26 27 mp1i N Fin R V m B x N , y N S x m y V
29 7 22 23 25 28 ovmpod N Fin R V N matToPolyMat R = m B x N , y N S x m y
30 1 29 syl5eq N Fin R V T = m B x N , y N S x m y