text: The equivalence of the two formulations can also be seen by a simple parity argument without appeal to case analysis.
  Proposition. The following formulations of the Woodward–Hoffmann rules are equivalent:
  (A) For a pericyclic reaction, if the sum of the number of suprafacial 4q + 2 components and antarafacial 4r components is odd then it is thermally allowed; otherwise the reaction is thermally forbidden.
  (B) For a pericyclic reaction, if the total number of antarafacial components of a (4n + 2)-electron reaction is even or the total number of antarafacial components of a 4n-electron reaction is odd then it is thermally allowed; otherwise the reaction is thermally forbidden.
  Proof of equivalence: Index the components of a k-component pericyclic reaction  and assign component i with Woodward-Hoffmann symbol N the electron count and topology parity symbol  according to the following rules:<blockquote><math>n_i = \begin{cases} 0, & N\equiv 0\ (\mathrm{mod}\ 4)\\ 1, & N\equiv 2\ (\mathrm{mod}\ 4) \end{cases}\quad\mathrm{and}\quad
  p_i = \begin{cases} 0, & i\text{ is supra}\\ 1, & i\text{ is antara} \end{cases}.We have a mathematically equivalent restatement of (A):
  (A) A collection of symbols  is thermally allowed if and only if the number of symbols with the property  is odd.'
  Since the total electron count is 4n + 2 or 4n precisely when  (the number of (4q + 2)-electron components) is odd or even, respectively, while  gives the number of antarafacial components, we can also restate (B):
  (B) A collection of symbols  is thermally allowed if and only if exactly one of  or  is odd.'
  It suffices to show that (A) and (B) are equivalent. Exactly one of  or  is odd if and only if  is odd.  If ,  holds; hence, omission of symbols with the property  from a collection will not change the parity of .  On the other hand, when , we have , but  simply enumerates the number of components with the property .  Therefore, <blockquote><math>\sum_i n_i+p_i\equiv\sum_{n_i\neq p_i} n_i+p_i=
  \sum_{n_i\neq p_i} 1=\#\{(n_i,p_i,i)\ |\ n_i\neq p_i\}\ 
  (\mathrm{mod}\ 2)Thus,  and the number of symbols in a collection with the property  have the same parity.  Since formulations (A) and (B) are equivalent, so are (A) and (B), as claimed.  
  To give a concrete example, a hypothetical reaction with the descriptor [6 + 4 + 2] would be assigned the collection {(1, 0, 1), (0, 1, 2), (1, 1, 3)} in the scheme above.  There are two components, (1, 0, 1) and (0, 1, 2), with the property , so the reaction is not allowed by (A).  Likewise,  and  are both even, so (B) yields the same conclusion (as it must): the reaction is not allowed.