theorem
  (minfuncreal(A)).((maxfuncreal(A)).(f,g),f)=f
proof
  thus (minfuncreal(A)).((maxfuncreal(A)).(f,g),f) =(minfuncreal(A)).((
  maxfuncreal(A)).(g,f),f) by Th8
    .=f by Th18;
end;
