theorem Th17:
  (minfuncreal(A)).(f,(maxfuncreal(A)).(g,f))=f
proof
  thus (minfuncreal(A)).(f,(maxfuncreal(A)).(g,f)) =(minfuncreal(A)).(f,(
  maxfuncreal(A)).(f,g)) by Th8
    .=f by Th16;
end;
