theorem Th69:
  TS is deterministic implies for P, Q being RedSequence of
  ==>.-relation(TS) st P.1 = Q.1 & (P.len P)`2 = (Q.len Q)`2 holds P = Q
proof
  assume
A1: TS is deterministic;
  then
A2: not <%>E in rng dom (the Tran of TS);
  let P, Q be RedSequence of ==>.-relation(TS) such that
A3: P.1 = Q.1 and
A4: (P.len P)`2 = (Q.len Q)`2;
  per cases by XXREAL_0:1;
  suppose
    len P = len Q;
    hence thesis by A1,A3,Th68;
  end;
  suppose
A5: len P > len Q;
    len P >= 0 + 1 by NAT_1:13;
    then
A6: len P in dom P by FINSEQ_3:25;
    set k = len Q;
A7: k >= 0 + 1 by NAT_1:13;
    then
A8: k in dom P by A5,FINSEQ_3:25;
    k in dom Q by A7,FINSEQ_3:25;
    then (P.len Q)`2 = (P.len P)`2 by A1,A3,A4,A8,Th67;
    hence thesis by A2,A5,A8,A6,Th66;
  end;
  suppose
A9: len P < len Q;
    len Q >= 0 + 1 by NAT_1:13;
    then
A10: len Q in dom Q by FINSEQ_3:25;
    set k = len P;
A11: k >= 0 + 1 by NAT_1:13;
    then
A12: k in dom Q by A9,FINSEQ_3:25;
    k in dom P by A11,FINSEQ_3:25;
    then (Q.len P)`2 = (Q.len Q)`2 by A1,A3,A4,A12,Th67;
    hence thesis by A2,A9,A12,A10,Th66;
  end;
end;
