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 Th61:
  width A <> 0 or len A = 0 implies Solutions_of(A,len A|->0.K) is non empty
proof
  set L=len A|->0.K;
A1: len L=len A by CARD_1:def 7;
  reconsider A9=A as Matrix of len A,width A,K by MATRIX_0:51;
  assume
A2: width A <> 0 or len A = 0;
  per cases by A2;
  suppose
    len A=0;
    then Solutions_of(A9,ColVec2Mx L) = {{}} by A1,Th51;
    then
A3: {} in Solutions_of(A9,ColVec2Mx L) by TARSKI:def 1;
    then consider f such that
A4: {} = ColVec2Mx f and
    len f = width A by Th58;
    f in Solutions_of(A,L) by A3,A4;
    hence thesis;
  end;
  suppose
A5: width A > 0;
    ColVec2Mx L =0.(K,len A,1) by Th32;
    then len ColVec2Mx L=len L & the_rank_of A=the_rank_of (A^^(ColVec2Mx L))
    by Th23,MATRIX_0:def 2;
    then Solutions_of(A,ColVec2Mx L) is non empty by A1,A5,Th57;
    then consider x being object such that
A6: x in Solutions_of(A,ColVec2Mx L);
    consider f such that
A7: x = ColVec2Mx f and
    len f = width A by A6,Th58;
    f in Solutions_of(A,L) by A6,A7;
    hence thesis;
  end;
end;
