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I am trying to define a recursive function

$g : DNF \rightarrow CNF$

such that for any propositional formula $\varphi$, $g(\varphi)\equiv \neg\varphi$.

I tried to split that into cases, such as follows:

  • If $\varphi=p$ then $g(\varphi)=\neg p$     where $p$ is atomic propositional variable
  • If $\varphi=\neg p$ then $g(\varphi)=p$     where $p$ is atomic propositional variable
  • If $\varphi=\alpha\vee\beta$, then $g(\varphi)=\neg g(\alpha)\wedge\neg g(\beta)$
  • If $\varphi=\alpha\wedge\beta$, then $g(\varphi)=\neg g(\alpha)\vee\neg g(\beta)$
  • ...

But for the complex binary connective I am not sure the result is actually a CNF form of $\neg \varphi$, nor I know how to prove it.

Please help me to define such a function and help me to prove it's correctness.

Please advise. Thank you.

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    $\begingroup$ Just try your proposal out on $(p \land q) \lor r$. You won't get a CNF because, in a CNF (or a DNF), negation is restricted to negation of atoms. To fix this, just delete the $\lnot$s in the clauses for $\lor$ and $\land$ (and get rid of the clause for ..., as there is no other possibility for a DNF input). $\endgroup$ – Rob Arthan Mar 27 at 12:55
  • $\begingroup$ Thanks @RobArthan, did you mean to delete the $\neg$ in the clauses for $\vee$ and $\wedge$? $\endgroup$ – Dennis Mar 27 at 12:57
  • $\begingroup$ Yes - I've fixed the comment. $\endgroup$ – Rob Arthan Mar 27 at 12:57
  • $\begingroup$ So you say that $g(p)=\neg p$ ; $g(\neg p)=p$ ; $g(\alpha\vee\beta)=g(\alpha)\wedge g(\beta)$ ; $g(\alpha\wedge\beta)=g(\alpha)\vee g(\beta)$ is a correct definition? Any idea how to justify it simply? Thank you very much! $\endgroup$ – Dennis Mar 27 at 13:03
  • $\begingroup$ I have written up an answer. $\endgroup$ – Rob Arthan Mar 27 at 13:20
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You can fix your definition just by removing the negations from the clauses for $\alpha \lor \beta$ and $\alpha \land \beta$. If you define: $$ \begin{align*} g(p) &= \lnot p \\ g(\lnot p) &= p\\ g(\alpha \lor \beta) &= g(\alpha) \land g(\beta) \\ g(\alpha \land \beta) &= g(\alpha) \lor g(\beta) \\ \end{align*} $$ where $p$ denotes an atom and $\alpha$ and $\beta$ are any formulas. Then (1) $g(\phi)$ is well-defined for any $\phi$ that is in DNF (because in a DNF negations only occur in literals) and (2) $g(\phi) \equiv \lnot \phi$, for any $\phi$ in DNF, which you can prove by induction on the size of the DNF formula $\phi$: it is clear in the base cases, when $\phi$ is $p$ or $\lnot p$, and in the inductive steps, if $g(\alpha) \equiv \lnot\alpha$ and $g(\beta) \equiv \lnot \beta$, then you have: $$ g(\alpha \lor \beta) = g(\alpha) \land g(\beta) \equiv \lnot \alpha \land \lnot\beta \equiv \lnot(\alpha \lor \beta) $$ and similarly for $g(\alpha \land \beta)$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Dennis Mar 27 at 13:39

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