User blog comment:Edwin Shade/The Most Powerful Array Notation Yet Devised/@comment-5529393-20170924012309/@comment-32876686-20170924024620

Ah yes, thank you for pointing out that typo, I spent about twenty minutes revising the final post after noting there were a few mistakes.

As for the definition of $$N(a)$$, I understand that a function with multiple outputs for the same input is not a function, so I will change it from 'the Nova function' accordingly to 'the Nova operator'. But other than that, the Nova operator works like I intended it to.

The language I will use to define ordinals and their fundamental sequences is in the making, and I'll admit the Nova operator may not be as powerful as I originally thought it to be, but I believed that $$F_{N(\omega,\omega)}(10^{100})$$ is bigger than BIG FOOT for the following line of reasoning.

According to the Wiki article, BIG FOOT is equal to $$FOOT^{10}(10^{100})$$, where recursion is being used. I took this to mean that $$F(n)$$ was equal to the highest number that can be made using arbitrarily high order set-theory, (as the page seemed to indicate), and that this operation was being recurred on a Googol ten times. The language I plan to use for $$N(n)$$ is arbitrarily high order set-theory, and so consequently I thought $$F_{N(\omega,\omega)}(10^{100})$$ to be bigger than BIGFOOT.

Looking back at the article though, I see if the growth rate of $$FOOT(n)$$ exceeds all nth-order set theory functions, then $$F_{N(\omega,\omega)}(10^{100})$$ is most likely not bigger than BIG FOOT.

However, seeing as 'First-Order Oodle Theory' is involved with BIGFOOT, and there probably is such as a thing as 'Second-Order Oodle Theory', I could just say that the limit of my language is nth-Order Oodle Theory. THEN $$F_{N(\omega,\omega)}(10^{100})$$ would probably be bigger than BIG FOOT, (although by means of a naive extension it would still probably be technically bigger).