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 Th54:
  n > 0 & k > 0 implies Solutions_of(0.(K,n,k),0.(K,n,m)) = 
  the set of all X where X is Matrix of k,m,K
proof
  assume that
A1: n > 0 and
A2: k > 0;
  set B=0.(K,n,m);
A3: width B=m by A1,MATRIX_0:23;
  set XX=the set of all X where X is Matrix of k,m,K;
  set A=0.(K,n,k);
  thus Solutions_of(A,B) c= XX
  proof
    let x be object;
    assume x in Solutions_of(A,B);
    then x is Matrix of k,m,K by A1,Th53;
    hence thesis;
  end;
  let x be object;
  assume x in XX;
  then consider X be Matrix of k,m,K such that
A4: x=X;::: and
  ::::not contradiction;
A5: width A=k & len X=k by A1,A2,MATRIX_0:23;
A6: width X=m by A2,MATRIX_0:23;
  len A=n by A1,MATRIX_0:23;
  then A*X=B by A1,A5,A6,MATRIX_5:22;
  hence thesis by A4,A3,A5,A6;
end;
