theorem
 dom g misses dom h implies f +* g +* h = f +* h +* g
proof assume
A1: dom g misses dom h;
 thus f +* g +* h = f +* (g +* h) by Th14
  .= f +* (h +* g) by A1,Th35
  .= f +* h +* g by Th14;
end;
