theorem
  j in dom f & j+1 in dom f implies LSeg(f^h,j) = LSeg(f,j)
proof
  assume that
A1: j in dom f and
A2: j+1 in dom f;
A3: 1 <= j & j+1 <= len f by A1,A2,FINSEQ_3:25;
  dom f c= dom(f^h) by FINSEQ_1:26;
  then
A4: j+1 <= len(f^h) by A2,FINSEQ_3:25;
  set p1 = f/.j, p2 = f/.(j+1);
A5: 1 <= j by A1,FINSEQ_3:25;
  p1 = (f^h)/.j & p2 = (f^h)/.(j+1) by A1,A2,FINSEQ_4:68;
  then LSeg(f^h,j) = LSeg(p1,p2) by A5,A4,TOPREAL1:def 3;
  hence thesis by A3,TOPREAL1:def 3;
end;
