theorem Th65:
  M is commutative associative
implies
  M $$ ([#]dom (F^G), A "**" (F^G)) =
    M.(M $$ ([#]dom F, A "**" F), M $$ ([#]dom G, A "**" G))
proof
A1: dom (A "**" (F^G) ) = dom (F^G)
    & dom (A "**" F)=dom F & dom (A "**" G)=dom G by FUNCT_2:def 1;
  A "**" (F^G) = (A "**" F) ^ (A "**" G) by Th63;
  hence thesis by A1,Th64;
end;
