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 Th62:
  not <%>E in rng dom (the Tran of TS) implies for P being
  RedSequence of ==>.-relation(TS) st P.1 = [x, u] & P.len P = [y, <%>E] holds
  len P <= len u + 1
proof
  defpred P[Nat] means for P being RedSequence of ==>.-relation(TS), u, x st
  len u = $1 & P.1 = [x, u] & P.len P = [y, <%>E] holds len P <= len u + 1;
  assume
A1: not <%>E in rng dom (the Tran of TS);
A2: now
    let k;
    assume
A3: for n st n < k holds P[n];
    now
      let P be RedSequence of ==>.-relation(TS), u, x such that
A4:   len u = k and
A5:   P.1 = [x, u] and
A6:   P.len P = [y, <%>E];
      per cases;
      suppose
        len u < 1;
        then u = <%>E by NAT_1:25;
        then len P = 1 by A1,A5,A6,Th60;
        hence len P <= len u + 1 by NAT_1:25;
      end;
      suppose
A7:     len u >= 1;
A8:     len P <> 1
        proof
          assume len P = 1;
          then u = <%>E by A5,A6,XTUPLE_0:1;
          hence contradiction by A7;
        end;
        then consider P9 being RedSequence of ==>.-relation(TS) such that
A9:     <*P.1*>^P9 = P and
A10:    len P9 + 1 = len P by Th5,NAT_1:25;
A11:    len P > 1 by A8,NAT_1:25;
        then len P >= 1 + 1 by NAT_1:13;
        then
A12:    1 + 1 in dom P by FINSEQ_3:25;
A13:    1 in dom P by A11,FINSEQ_3:25;
        then
A14:    P.(1 + 1) = [(P.(1 + 1))`1, (P.(1 + 1))`2] by A12,Th48;
        then
A15:    [[x, u], [(P.(1 + 1))`1, (P.(1 + 1))`2]] in ==>.-relation(TS) by A5,A13
,A12,REWRITE1:def 2;
        then reconsider u1 = (P.(1 + 1))`2 as Element of E^omega by Th32;
A16:    len <*P.1*> = 1 & len P9 >= 1 by FINSEQ_1:39,NAT_1:25;
        then
A17:    P9.1 = [(P.(1 + 1))`1, u1] by A9,A14,FINSEQ_1:65;
        x, u ==>. (P.(1 + 1))`1, u1, TS by A15,Def4;
        then consider v such that
A18:    x, v -->. (P.(1 + 1))`1, TS & u = v^u1;
        v <> <%>E & len u = len u1 + len v by A1,A18,Th15,AFINSQ_1:17;
        then
A19:    len u - 0 > len u1 + len v - len v by XREAL_1:15;
        then
A20:    len u1 + 1 <= len u by NAT_1:13;
        P9.len P9 = [y, <%>E] by A6,A9,A10,A16,FINSEQ_1:65;
        then len P9 <= len u1 + 1 by A3,A4,A19,A17;
        then len P9 <= len u by A20,XXREAL_0:2;
        hence len P <= len u + 1 by A10,XREAL_1:6;
      end;
    end;
    hence P[k];
  end;
  for k holds P[k] from NAT_1:sch 4(A2);
  hence thesis;
end;
