rker &quot;x&quot; with a stick stemming from the site of &quot;x&quot;.

5.  In the case of an expression Q that does not bear a token of the
    designated marker &quot;x&quot;, it is stipulated that Q[o/x] = Q = Q[|/x].

In the case of e_1 and e_2, we have:

    e_1  =  e_3 [o/x]

    e_2  =  e_3 [|/x]

We've already noted that [e_1] = [e_3 [o/x]] = [!e!] = [e_3 [|/x]] = [e_2].

Jon Awbrey

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

</PRE>

<!--X-Body-of-Message-End-->
<!--X-MsgBody-End-->
<!--