theorem Th29:
  {x} common_on_dom H implies (r(#)H)#x = r(#)(H#x)
proof
  assume
A1: {x} common_on_dom H;
  now
    let n be Element of NAT;
    x in {x} & {x} c= dom(H.n) by A1,TARSKI:def 1;
    then x in dom (H.n);
    then
A2: x in dom (r(#)(H.n)) by VALUED_1:def 5;
    thus ((r(#)H)#x).n = ((r(#)H).n).x by Def10
      .= (r(#)(H.n)).x by Def1
      .= r*((H.n).x) by A2,VALUED_1:def 5
      .= r*((H#x).n) by Def10
      .= (r(#)(H#x)).n by SEQ_1:9;
  end;
  hence thesis by FUNCT_2:63;
end;
