theorem
  f/.1 = E-min L~f & E-min L~f <> S-max L~f implies (E-min L~f)..f < (
  S-max L~f)..f
proof
  assume that
A1: f/.1 = E-min L~f and
A2: E-min L~f <> S-max L~f;
A3: S-max L~f in rng f by SPRECT_2:42;
  then (S-max L~f)..f in dom f by FINSEQ_4:20;
  then
A4: (S-max L~f)..f >= 1 by FINSEQ_3:25;
  E-min L~f in rng f & (E-min L~f)..f = 1 by A1,FINSEQ_6:43,SPRECT_2:45;
  hence thesis by A3,A2,A4,FINSEQ_5:9,XXREAL_0:1;
end;
