theorem Th4:
  f c= g implies f\h c= g\h
proof
  assume
A1: f.c <= g.c;
  let c;
  f.c <= g.c by A1;
  then min(f.c,(1_minus h).c) <= min(g.c,(1_minus h).c) by XXREAL_0:18;
  then (f\h).c <= min(g.c,(1_minus h).c) by FUZZY_1:5;
  hence thesis by FUZZY_1:5;
end;
