Editing Algebraic normal form (section)

== Performing operations within algebraic normal form ==
There are straightforward ways to perform the standard Boolean operations on ANF inputs in order to get ANF results.

XOR (logical exclusive disjunction) is performed directly:
: ({{fontcolor|red|1 ⊕ x}}) ⊕ ({{fontcolor|green|1 ⊕ x ⊕ y}})
: {{fontcolor|red|1 ⊕ x}} ⊕ {{fontcolor|green|1 ⊕ x ⊕ y}}
: 1 ⊕ 1 ⊕ x ⊕ x ⊕ y
: y

NOT (logical negation) is XORing 1:<ref name="not-equiv">[http://www.wolframalpha.com/input/?i=simplify+1+xor+a WolframAlpha NOT-equivalence demonstration: ¬a = 1 ⊕ a]</ref>
: {{fontcolor|red|¬}}{{fontcolor|green|(1 ⊕ x ⊕ y)}}
: {{fontcolor|red|1 ⊕ }}{{fontcolor|green|(1 ⊕ x ⊕ y)}}
: 1 ⊕ 1 ⊕ x ⊕ y
: x ⊕ y

AND (logical conjunction) is [[distributive property|distributed algebraically]]<ref name="and-equiv">[http://www.wolframalpha.com/input/?i=%28a+xor+b%29+and+%28c+xor+d%29+in+anf WolframAlpha AND-equivalence demonstration: (a ⊕ b)(c ⊕ d) = ac ⊕ ad ⊕ bc ⊕ bd]</ref>
: ({{fontcolor|red|1}} ⊕ {{fontcolor|red|x}}){{fontcolor|green|(1 ⊕ x ⊕ y)}}
: {{fontcolor|red|1}}{{fontcolor|green|(1 ⊕ x ⊕ y)}} ⊕ {{fontcolor|red|x}}{{fontcolor|green|(1 ⊕ x ⊕ y)}}
: (1 ⊕ x ⊕ y) ⊕ (x ⊕ x ⊕ xy)
: 1 ⊕ x ⊕ x ⊕ x ⊕ y ⊕ xy
: 1 ⊕ x ⊕ y ⊕ xy

OR (logical disjunction) uses either 1 ⊕ (1 ⊕ a)(1 ⊕ b)<ref name="or-demorgans">From [[De Morgan's laws]]</ref> (easier when both operands have purely true terms) or a ⊕ b ⊕ ab<ref name="or-equiv">[http://www.wolframalpha.com/input/?i=simplify+a+xor+b+xor+%28a+and+b%29 WolframAlpha OR-equivalence demonstration: a + b = a ⊕ b ⊕ ab]</ref> (easier otherwise):
: ({{fontcolor|red|1 ⊕ x}}) + ({{fontcolor|green|1 ⊕ x ⊕ y}})
: 1 ⊕ (1 ⊕ {{fontcolor|red|1 ⊕ x}})(1 ⊕ {{fontcolor|green|1 ⊕ x ⊕ y}})
: 1 ⊕ x(x ⊕ y)
: 1 ⊕ x ⊕ xy