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