reserve i,j,k,l,m,n for Nat,
  D for non empty set,
  f for FinSequence of D;

theorem Th2:
  i > j & i in dom f & j in dom f & k in dom mid(f,i,j) implies i
  -' k + 1 in dom f
proof
  assume that
A1: i > j and
A2: i in dom f and
A3: j in dom f;
A4: i <= len f by A2,FINSEQ_3:25;
A5: 1+0 <= j by A3,FINSEQ_3:25;
  then 1 - j <= 0 by XREAL_1:47;
  then
A6: i +(1 - j) <= i + 0 by XREAL_1:6;
  assume
A7: k in dom mid(f,i,j);
  then
A8: k <= len mid(f,i,j) by FINSEQ_3:25;
  k >= 1 by A7,FINSEQ_3:25;
  then 1 - k <= 0 by XREAL_1:47;
  then i + (1 - k) <= i+0 by XREAL_1:6;
  then
A9: i - k + 1 <= i;
  1+0 <= i & j <= len f by A2,A3,FINSEQ_3:25;
  then len mid(f,i,j) = i -' j +1 by A1,A4,A5,FINSEQ_6:118;
  then k <= i - j +1 by A1,A8,XREAL_1:233;
  then i -' k + 1 <= i by A6,A9,XREAL_1:233,XXREAL_0:2;
  then
A10: i -' k + 1 <= len f by A4,XXREAL_0:2;
  1 <= i -' k + 1 by NAT_1:11;
  hence thesis by A10,FINSEQ_3:25;
end;
