theorem Th18:
  (minfuncreal(A)).((maxfuncreal(A)).(g,f),f)=f
proof
  thus (minfuncreal(A)).((maxfuncreal(A)).(g,f),f) =(minfuncreal(A)).(f,(
  maxfuncreal(A)).(g,f)) by Th9
    .=f by Th17;
end;
