theorem Th60:
  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, u] holds len P
  = 1 & x = y
proof
  defpred P[Nat] means for P being RedSequence of ==>.-relation(TS), x, y st P
  .1 = [x, u] & P.len P = [y, u] & len P = $1 & $1 <> 1 holds contradiction;
  assume
A1: not <%>E in rng dom (the Tran of TS);
A2: now
    let k;
    assume P[k];
    now
      let P be RedSequence of ==>.-relation(TS), x, y such that
A3:   P.1 = [x, u] and
A4:   P.len P = [y, u] and
A5:   len P = k + 1 & k + 1 <> 1;
      consider Q being RedSequence of ==>.-relation(TS) such that
      <*P.1*>^Q = P and
A6:   len P = len Q + 1 by A5,Th5,NAT_1:25;
      len Q >= 0 + 1 by NAT_1:8;
      then len Q + 1 >= 1 + 1 by XREAL_1:6;
      then
A7:   1 + 1 in dom P by A6,FINSEQ_3:25;
      len P > 1 by A5,NAT_1:25;
      then
A8:   1 in dom P by FINSEQ_3:25;
      then P.(1 + 1) = [(P.(1 + 1))`1, (P.(1 + 1))`2] by A7,Th48
        .= [(P.(1 + 1))`1, u] by A3,A4,A7,Th54;
      then [[x, u], [(P.(1 + 1))`1, u]] in ==>.-relation(TS) by A3,A8,A7,
REWRITE1:def 2;
      then x, u ==>. (P.(1 + 1))`1, u, TS by Def4;
      hence contradiction by A1,Th28;
    end;
    hence P[k + 1];
  end;
  let P be RedSequence of ==>.-relation(TS) such that
A9: P.1 = [x, u] & P.len P = [y, u];
A10: P[0];
  for k holds P[k] from NAT_1:sch 2(A10, A2);
  hence len P = 1 by A9;
  hence thesis by A9,XTUPLE_0:1;
end;
