 De Morgan dual of a Boolean function

For a Boolean function $f$ of $n$ input bits, define its De Morgan dual by complementing every input, applying $f$, and complementing the output:

De Morgan's laws say that the De Morgan dual of logical conjunction is logical disjunction, and conversely. The constant zero Boolean function becomes constant one. This is different from merely negating the output: for AND, output logical negation gives one at $(0,0)$, whereas the De Morgan dual gives zero there.

Taking the De Morgan dual twice restores the original Boolean function:Here each input is complemented twice as well. For a Boolean formula built from AND, OR, NOT and constants, its De Morgan dual can therefore be obtained by exchanging AND with OR and zero with one throughout, leaving input names and NOT in place. A structural induction proves the rule: it holds for an input and each constant, De Morgan's laws handle the two binary logical connectives, and double logical negation handles NOT. This symmetry lets one translate a complex test systematically instead of guessing where its failures occur.