theorem
  (f1<++>f2) <##> f = (f1<##>f) <++> (f2<##>f)
proof
  set f3 = f1<##>f, f4 = f2<##>f, f5 = f1<++>f2;
A1: f1<##>f = f<##>f1 & f2<##>f = f<##>f2 by Th83;
  thus f5 <##> f = f <##> f5 by Th83
    .= f3 <++> f4 by A1,Th87;
end;
