theorem Th20:
  f is one-to-one & 1 <= i & i+1 <= len f implies LSeg(f,i) is non trivial
proof
  assume
A1: f is one-to-one;
A2: i <> i+1;
  assume
A3: 1 <= i & i+1 <= len f;
  then i in dom f & i+1 in dom f by SEQ_4:134;
  then
A4: f/.i<>f/.(i+1) by A1,A2,PARTFUN2:10;
A5: f/.i in LSeg(f/.i,f/.(i+1)) & f/.(i+1) in LSeg(f/.i,f/.(i+1)) by
RLTOPSP1:68;
  LSeg(f/.i,f/.(i+1)) = LSeg(f,i) by A3,TOPREAL1:def 3;
  hence thesis by A4,A5,ZFMISC_1:def 10;
end;
