theorem
  (maxfuncreal(A)).(f,(minfuncreal(A)).(g,f))=f
proof
  thus (maxfuncreal(A)).(f,(minfuncreal(A)).(g,f)) =(maxfuncreal(A)).(f,(
  minfuncreal(A)).(f,g)) by Th9
    .=f by Th12;
end;
