a formula is already in pure ANF, so you only need to perform the formula's boolean operations as shown above to get the entire formula into ANF. For example: x + (y ⋅ ¬z) x + (y(1 ⊕ z)) x + (y ⊕ yz) x ⊕ (y ⊕ yz) ⊕ x(y ⊕ yz) x ⊕ y ⊕ xy ⊕ yz ⊕ xyz Formal representation ANF is sometimes described in an equivalent way: {| cellpadding="4" |- | | |- | | |- | | |- | | |- | | |} where fully describes . Recursively deriving multiargument Boolean functions There are only four functions with one argument: To represent a function with multiple arguments one can use the following equality: , where Indeed, if then and so if then and so Since both and have fewer arguments than it follows that using this process recursively we will finish with functions with one variable. For example, let us construct ANF of (logical or): since and it follows that by distribution, we get the final ANF: See also Reed–Muller expansion Zhegalkin normal form Boolean function Logical graph Zhegalkin polynomial Negation normal form Conjunctive normal form Disjunctive normal form Karnaugh map Boolean ring References Further reading Boolean algebra Normal forms (logic) ru:Полином Жегалкина