        @              |
|                                                           |
|       (p (q)) (p (r))       =          (p  (q r))         |
|                                                           |
|       [p=&gt;q] &amp; [p=&gt;r]       =          [p=&gt;[q&amp;r]]         |
|                                                           |
o-----------------------------------------------------------o

I know that it must seem tedious, but I probably ought to go ahead and
carry out the second half 