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I'm trying to understand the relations between $\Delta$-formulas and $\Sigma$-formulas and $\Pi$-formulas.

As far as I know, if all quantifiers of a formula are bounded, then the formula is $\Delta$-formula; And being a $\Delta$-formula leads to being both $\Sigma$-formula and $\Pi$-formula. Because $\Delta$-formulas are actually the intersection of $\Sigma$-formulas and $\Pi$-formulas.

Now Let's consider following formula:

$\phi :\equiv (x

As there is no unbounded quantifier, then $\phi$ is a $\Delta$-formula (i.e., $\phi$ is both $\Sigma$-formula and $\Pi$-formula). Furthermore,

$\neg \phi :\equiv \neg(x

which is a $\Delta$-formula, as well.

Can I generally claim the following argument?:

If $\phi$ is a $\Delta$-formula, then $\neg \phi$ is a $\Delta$-formula, too

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Looks like you can. By hypothesis, $\phi$ is of the form $\forall x (x<\overline{n} \land \varphi(x))$ or $\exists x (x<\overline{n} \rightarrow \varphi(x))$ (up to logical equivalence). Apply negation and play around with the symbols a bit and you'll find $\neg \phi$ can be expressed (up to logical equivalence) in a similar format.

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    Thanks. So In other words, I can say that a $\Delta$-formula and its negation are both simultaneously $\Delta$-formula, $\Sigma$-formula and $\Pi$-formula, Right? – 2017-02-27
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    Yeah...and more generally the negation of a $\Sigma$ formula is a $\Pi$ formula and vice versa (literally by De Morgan's law). Like you said, a $\Delta$ formula is simply the intersection, which is just the definition of most references. But there's an important intuition to that, though: bounded quantifiers are not REALLY quantifiers since they're logically equivalent to a finite conjunction/disjunction of instances of the inner formula. – 2017-02-27
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    I'm all clear now. You helped a lot. Much appreciated! – 2017-02-27
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    Glad I can help. You still might appreciate my edits of the last comment. – 2017-02-27