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);

theorem
  for A,B be Matrix of K st len A = len B & len A = the_rank_of A holds
  the_rank_of A = the_rank_of (A^^B)
proof
  let A,B be Matrix of K such that
A1: len A = len B and
A2: len A =the_rank_of A;
  set L=len A;
  reconsider B9=B as Matrix of L,width B,K by A1,MATRIX_0:51;
  reconsider A9=A as Matrix of L,width A,K by MATRIX_0:51;
A3: the_rank_of (A9^^B9)<=len (A9^^B9) & len (A9^^B9)=L by MATRIX13:74
,MATRIX_0:def 2;
  the_rank_of (A9^^B9)>=L by A1,A2,Th20;
  hence thesis by A2,A3,XXREAL_0:1;
end;
