theorem Th36:
  for V be finite-rank free Z_Module
  holds for A, B being Basis of V holds
  card A = card B
  proof
    let V be finite-rank free Z_Module;
    let A, B be Basis of V;
    set p = the prime Element of INT.Ring;
    set AQ = {ZMtoMQV(V, p, u) where u is Vector of V : u in A};
    now let x be object;
      assume x in AQ; then
      ex u be Vector of V st x = ZMtoMQV(V, p, u) & u in A;
      hence x in the carrier of Z_MQ_VectSp(V, p);
    end;
    then reconsider AQ as Subset of Z_MQ_VectSp(V, p) by TARSKI:def 3;
    set BQ = {ZMtoMQV(V, p, u) where u is Vector of V : u in B};
    now let x be object;
      assume x in BQ;
      then ex u be Vector of V st x = ZMtoMQV(V, p, u) & u in B;
      hence x in the carrier of Z_MQ_VectSp(V, p);
    end;
    then reconsider BQ as Subset of Z_MQ_VectSp(V, p) by TARSKI:def 3;
    A1: card A = card AQ by Th26;
    A2: card B = card BQ by Th26;
    A3: AQ is Basis of Z_MQ_VectSp(V, p) by Th35;
    BQ is Basis of Z_MQ_VectSp(V, p) by Th35;
    hence card A = card B by A1,A2,A3,VECTSP_9:22;
  end;
