let k, n be Nat; :: thesis: for K being Field
for B being Matrix of K st not Solutions_of ((0. (K,n,k)),B) is empty holds
B = 0. (K,n,())

let K be Field; :: thesis: for B being Matrix of K st not Solutions_of ((0. (K,n,k)),B) is empty holds
B = 0. (K,n,())

let B be Matrix of K; :: thesis: ( not Solutions_of ((0. (K,n,k)),B) is empty implies B = 0. (K,n,()) )
set A = 0. (K,n,k);
set ZERO = 0. (K,n,());
assume not Solutions_of ((0. (K,n,k)),B) is empty ; :: thesis: B = 0. (K,n,())
then consider x being object such that
A1: x in Solutions_of ((0. (K,n,k)),B) ;
A2: len (0. (K,n,k)) = n by MATRIX_0:def 2;
A3: dom (0. (K,n,k)) = Seg n ;
A4: len (0. (K,n,())) = n by MATRIX_0:def 2;
then A5: len B = len (0. (K,n,())) by A1, A2, Th33;
then reconsider B9 = B as Matrix of n, width B,K by ;
A6: ex X being Matrix of K st
( X = x & len X = width (0. (K,n,k)) & width X = width B & (0. (K,n,k)) * X = B ) by A1;
now :: thesis: for i being Nat st 1 <= i & i <= n holds
B . i = (0. (K,n,())) . i
let i be Nat; :: thesis: ( 1 <= i & i <= n implies B . i = (0. (K,n,())) . i )
assume A7: ( 1 <= i & i <= n ) ; :: thesis: B . i = (0. (K,n,())) . i
A8: width (0. (K,n,k)) = k by ;
A9: i in Seg n by A7;
then Line ((0. (K,n,k)),i) = (0. (K,n,k)) . i by MATRIX_0:52
.= (width (0. (K,n,k))) |-> (0. K) by ;
then () |-> (0. K) = Line (B,i) by A1, A6, A3, A9, Th41
.= B9 . i by ;
hence B . i = (0. (K,n,())) . i by ; :: thesis: verum
end;
hence B = 0. (K,n,()) by A4, A5; :: thesis: verum