## FANDOM

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The "operator" that I'm going to show you is not an operator.

## Rules

Let O be an operator.

$>^n$ is not an operator.

$(lOm)>^0n=(lOm)Om$

$(lOm)>^n1=lOm$

$(kOl)>^mn=((kOl)>^mn-1)>^{m-1}(kOl)$

Lastly, we evaluate the $lOm$ if it is the innermost expression in that form that is directly after a $>^n$ or if there are no more $>^n$ signs in the expression.