reserve x,y for set,
  i,j,k,l,m,n for Nat,
  K for Field,
  N for without_zero finite Subset of NAT,
  a,b for Element of K,
  A,B,B1,B2,X,X1,X2 for (Matrix of K),
  A9 for (Matrix of m,n,K),
  B9 for (Matrix of m,k,K);
reserve D for non empty set,
  bD for FinSequence of D,
  b,f,g for FinSequence of K,
  MD for Matrix of D;

theorem
  n>0 & Solutions_of(0.(K,n,0),0.(K,n,m)) is non empty implies m = 0
proof
  assume that
A1: n>0 and
A2: Solutions_of(0.(K,n,0),0.(K,n,m)) is non empty;
  consider x being object such that
A3: x in Solutions_of(0.(K,n,0),0.(K,n,m)) by A2;
A4: width 0.(K,n,0)=0 by A1,MATRIX_0:23;
  ex X st X=x & len X = width 0.(K,n,0) & width X = width 0.(K,n,m) & 0.(K,
  n,0) * X = 0.(K,n,m) by A3;
  hence 0 = width 0.(K,n,m) by A4,MATRIX_0:def 3
    .= m by A1,MATRIX_0:23;
end;
