let D be non empty set ; for B being Matrix of 3,1,D holds B = <*<*(B * (1,1))*>,<*(B * (2,1))*>,<*(B * (3,1))*>*>
let B be Matrix of 3,1,D; B = <*<*(B * (1,1))*>,<*(B * (2,1))*>,<*(B * (3,1))*>*>
reconsider C = <*<*(B * (1,1))*>,<*(B * (2,1))*>,<*(B * (3,1))*>*> as Matrix of 3,1,D by Th25;
A1:
( len B = 3 & width B = 1 )
by MATRIX_0:23;
A2:
for i, j being Nat st [i,j] in Indices B holds
B * (i,j) = C * (i,j)
proof
let i,
j be
Nat;
( [i,j] in Indices B implies B * (i,j) = C * (i,j) )
A3:
Indices C = [:(Seg 3),(Seg 1):]
by MATRIX_0:23;
A4:
Indices B = [:(Seg 3),(Seg 1):]
by MATRIX_0:23;
assume A5:
[i,j] in Indices B
;
B * (i,j) = C * (i,j)
then
(
i in Seg 3 &
j in Seg 1 )
by A4, ZFMISC_1:87;
then
( (
i = 1 or
i = 2 or
i = 3 ) &
j = 1 )
by FINSEQ_1:2, FINSEQ_3:1, ENUMSET1:def 1, TARSKI:def 1;
end;
( len C = 3 & width C = 1 )
by MATRIX_0:23;
hence
B = <*<*(B * (1,1))*>,<*(B * (2,1))*>,<*(B * (3,1))*>*>
by A1, A2, MATRIX_0:21; verum