theorem Th44:
  {f}++{g,h} = {f+g,f+h}
proof
  thus {f}++{g,h} = {f}++({g}\/{h}) by ENUMSET1:1
    .= ({f}++{g}) \/ ({f}++{h}) by Th41
    .= {f+g} \/ ({f}++{h}) by Th43
    .= {f+g} \/ {f+h} by Th43
    .= {f+g,f+h} by ENUMSET1:1;
end;
