
theorem Th6:
for D be set, Y be FinSequenceSet of D,
    F be FinSequence of Y, m1,m2,k1,k2 be Nat
 st 1 <= m1 & 1 <= m2
  & m1 + Sum Length(F|k1) = m2 + Sum Length(F|k2)
  & m1 + Sum Length(F|k1) <= Sum Length(F|(k1+1))
  & m2 + Sum Length(F|k2) <= Sum Length(F|(k2+1))
 holds m1=m2 & k1=k2
proof
   let D be set, Y be FinSequenceSet of D, F be FinSequence of Y,
       m1,m2,k1,k2 be Nat;
   assume that
A1: 1 <= m1 & 1 <= m2 and
A2: m1 + Sum Length(F|k1) = m2 + Sum Length(F|k2) and
A3: m1 + Sum Length(F|k1) <= Sum Length(F|(k1+1)) and
A4: m2 + Sum Length(F|k2) <= Sum Length(F|(k2+1));
   set n = m1 + Sum Length(F|k1);
A5:now assume A6: k1 <> k2;
    per cases by A6,XXREAL_0:1;
    suppose k1 < k2; then
     k1+1 <= k2 by NAT_1:13; then
     Sum Length(F|(k1+1)) <= Sum Length(F|k2) by Th5; then
     n <= Sum Length(F|k2) by A3,XXREAL_0:2;
     hence contradiction by A2,A1,NAT_1:19;
    end;
    suppose k1 > k2; then
     k2+1 <= k1 by NAT_1:13; then
     Sum Length(F|(k2+1)) <= Sum Length(F|k1) by Th5; then
     n <= Sum Length(F|k1) by A2,A4,XXREAL_0:2;
     hence contradiction by A1,NAT_1:19;
    end;
   end;
   now assume m1 <> m2; then
    Sum Length(F|k1) - Sum Length(F|k2) <> 0 by A2;
    hence k1 <> k2;
   end;
   hence thesis by A5;
end;
