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

theorem Th1:
  i <= j & i in dom f & j in dom f & k in dom mid(f,i,j) implies k+
  i-'1 in dom f
proof
  assume that
A1: i <= j and
A2: i in dom f and
A3: j in dom f;
A4: j <= len f by A3,FINSEQ_3:25;
A5: 1+0 <= i by A2,FINSEQ_3:25;
  then i-1 >= 0 by XREAL_1:19;
  then
A6: k+0 <= k+(i-1) 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;
  i <= len f & 1 <= j by A2,A3,FINSEQ_3:25;
  then len mid(f,i,j) = j -' i +1 by A1,A5,A4,FINSEQ_6:118;
  then k <= j - i +1 by A1,A8,XREAL_1:233;
  then k <= j +1 - i;
  then k+i <= j +1 by XREAL_1:19;
  then k+i >= i & k+i - 1 <= j by NAT_1:11,XREAL_1:20;
  then k+i -' 1 <= j by A5,XREAL_1:233,XXREAL_0:2;
  then
A9: k+i-'1 <= len f by A4,XXREAL_0:2;
  1 <= k by A7,FINSEQ_3:25;
  then 1 <= k+i-1 by A6,XXREAL_0:2;
  then 1 <= k+i-'1 by NAT_D:39;
  hence thesis by A9,FINSEQ_3:25;
end;
