Forum:Rule in phi function

We know that, in phi function exist some ordinals in form \(\phi(\#\,\alpha,0,\cdots,0,\beta+1)\) (\(\alpha\) is a limit ordinal, \(\beta\) can be not). There are two variants for handling this:


 * \(\phi(\#\,\alpha,0,\cdots,0,\beta+1) = \phi(\#\,\alpha[n],\phi(\#\,\alpha,0,\cdots,0,\beta)+1,\cdots,0,\beta)\)


 * \(\phi(\#\,\alpha,0,\cdots,0,\beta+1) = \phi(\#\,\alpha[n],0,\cdots,\phi(\#\,\alpha,0,\cdots,0,\beta)+1,0)\)

So, must we need to place the successor of original expression where the last entry is reduced by 1 to the start or the end of the row of zeroes? Ikosarakt1 (talk ^ contribs) 10:45, July 12, 2013 (UTC)