theorem Th19:
  g (#) h /" k = g (#) (h /" k)
proof
A1: dom(g (#) (h /" k)) = dom g /\ dom(h /" k) & dom(g (#) h /" k) = dom(g
  (#) h ) /\ dom k by VALUED_1:16,def 4;
  dom(g (#) h) = dom g /\ dom h & dom(h /" k) = dom h /\ dom k by VALUED_1:16
,def 4;
  hence dom(g (#) h /" k) = dom(g (#) (h /" k)) by A1,XBOOLE_1:16;
  let x be object;
  assume x in dom(g (#) h /" k);
  thus (g (#) h /" k).x = (g (#) h).x / k.x by VALUED_1:17
    .= g.x * h.x / k.x by VALUED_1:5
    .= g.x * (h.x / k.x)
    .= g.x * (h /" k).x by VALUED_1:17
    .= (g (#) (h /" k)).x by VALUED_1:5;
end;
