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I'm reading about quantified boolean formulas. One sentence mentions:

You should also verify that the negation of the formula $Q_1x_1\cdot\cdot\cdot Q_nx_n \phi(x_1, ..., x_n)$ is the same as $Q^{\prime}_1x_1\cdot\cdot\cdot Q_n^{\prime}x_n \neg\phi(x_1, ..., x_n)$, where $Q^{\prime}_i$ is $\exists$ if $Q_i$ is $\forall$ and vice-versa.

I know the identity $\neg\forall\phi(x) = \exists x\neg\phi(x)$. Also, it seems like $\neg$ is, in some cases, distributive.

However, I haven't had any luck in using those two things to verify that statement.

Any idea?

Thanks!

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1 Answer 1

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You're right in that the identity you need is essentially $\neg \forall X. \phi = \exists X. \neg \phi$ and (symmetrically) $\neg \exists X. \phi = \forall X. \neg \phi$.

One can proceed via proof by induction.

Base case: It is obvious that $\neg Q_1 x_1. \phi(x_1)= Q_1' x_1. \neg \phi(x_1)$ via the properties above.

Inductive case: Assume the statement is true for $n-1$, i.e. $Q_1 x_1. Q_2 x_2., ..., Q_{n-1} x_{n-1}. \phi(x_1,...,x_{n-1}) = Q'_1 x_1... Q'_{n-1} x_{n-1}. \neg\phi(x_1,...,x_{n-1})$. Then

$$ Q_1 x_1,..., Q_n x_n. \phi_n (x_1,...,x_n)= Q_1x_1,...,Q_{n-1}x_{n-1} (Q_n x_n.\phi(x_1,...,x_n)) \\ = Q'_1 x_1...Q'_{n-1}x_{n-1}.\neg(Q_nx_n.\phi(x_1,...,x_n)) \text{ by ind.hyp.} \\ = Q'_1 x_1,...,Q'_nx_n.\neg\phi(x_1,...,x_n)\text{ by negation properties above.} $$

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    $\begingroup$ The reasoning is OK, however you are missing negations before the original quantifiers: $\lnot Q \phi = Q' \lnot \phi$. Also, the base case can even be where $n=0$, so without quantifiers. (The inductive case also works for $n=1$.) $\endgroup$ Commented May 31 at 21:44
  • $\begingroup$ @HendrikJan I corrected the typo in the base case. Thanks for pointing it out. $\endgroup$ Commented Jun 2 at 1:15
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    $\begingroup$ Sorry that I was not really explicit, but also the inductive case at the start needs a negation before the sequence of original quantifiers. $\endgroup$ Commented Jun 2 at 12:08

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