reserve x, x1, x2, y, y1, y2, z, z1, z2 for object, X, X1, X2 for set;
reserve E for non empty set;
reserve e for Element of E;
reserve u, u9, u1, u2, v, v1, v2, w, w1, w2 for Element of E^omega;
reserve F, F1, F2 for Subset of E^omega;
reserve i, k, l, n for Nat;
reserve TS for non empty transition-system over F;
reserve s, s9, s1, s2, t, t1, t2 for Element of TS;
reserve S for Subset of TS;

theorem Th67:
  TS is deterministic implies for P, Q being RedSequence of
==>.-relation(TS) st P.1 = Q.1 holds for k st k in dom P & k in dom Q holds P.k
  = Q.k
proof
  assume
A1: TS is deterministic;
  let P, Q be RedSequence of ==>.-relation(TS) such that
A2: P.1 = Q.1;
  defpred P[Nat] means $1 in dom P & $1 in dom Q implies P.$1 = Q.$1;
A3: now
    let k;
    assume
A4: P[k];
    now
      assume
A5:   k + 1 in dom P & k + 1 in dom Q;
      per cases;
      suppose
A6:     k in dom P & k in dom Q;
        then [P.k, P.(k + 1)] in ==>.-relation(TS) & [Q.k, Q.(k + 1)] in
        ==>.-relation(TS ) by A5,REWRITE1:def 2;
        hence P.(k + 1) = Q.(k + 1) by A1,A4,A6,Th44;
      end;
      suppose
        not k in dom P or not k in dom Q;
        then k = 0 by A5,REWRITE2:1;
        hence P.(k + 1) = Q.(k + 1) by A2;
      end;
    end;
    hence P[k + 1];
  end;
A7: P[0] by FINSEQ_3:25;
  for k holds P[k] from NAT_1:sch 2(A7, A3);
  hence thesis;
end;
