0
$\begingroup$

I came across this in my text book-

The negation of a statement of the form "$\exists x \in D$ such that $Q(x)$" is logically equivalent to a statement of the form "$\forall x \in D$ such that $\neg Q(x)$".

Symbolically, $\neg\big(\exists x\in D:Q(x)\big) ~\equiv~ \big(\forall x \in D:\neg Q(x)\big)$.

Negations are defined for statements.  Thus applying it on a propositional function is undefined as a propositional function is not a statement.  Here it shouldn't have been applied on $Q(x)$.

I have gone through Rosen's book and Wikipedia but they do not justify the usage of negation in this context, what am I missing?  I think the authors do not technically mean negation but that they mean something like the negation.

$\endgroup$
  • 1
    $\begingroup$ Rosen's book..... it's pretty not a good source learning logic from... I was a victim by learning logic by that book. I must say, there're many things not "correct" enough. And for a beginner, it's hard for he or she to sense this. $\endgroup$ – Eric Aug 16 '17 at 5:41
  • $\begingroup$ Rosen has simply omitted the fact that it is allowed to "produce" complex propositional functions from atomic ones: $P(x), Q(x), \ldots$, using boolean conncetives: $P(x) \to Q(x), \lnot Q(x), \ldots$. $\endgroup$ – Mauro ALLEGRANZA Aug 28 '17 at 15:07
2
$\begingroup$

You are not correct that negation applies only to statements (well-formed formulas without free variables). In fact, negation, like conjunction and disjunction, apply to any well-formed formulas, and any $n$-ary predicate symbol applied to $n$ variable letters or constant symbols is a well-formed formula.

In other words: $Q(x)$ is a well-formed formula; therefore $\lnot Q(x)$ is, too.

All this is explained on Wikipedia.

$\endgroup$
  • $\begingroup$ But the following article clearly states that the negation is applied to a proposition?link $\endgroup$ – magenn Aug 16 '17 at 5:51
  • 3
    $\begingroup$ @magenn: in predicate logic, $\lnot$ is an operator on well-formed formulas, and well-formed formulas include expressions such as $Q(x)$. If you think $\lnot$ shouldn't apply to $Q(x)$, you should also be complaining that $\land$ doesn't apply to $P(x)\land Q(x)$. But of course it does. $\endgroup$ – symplectomorphic Aug 16 '17 at 5:55
  • $\begingroup$ @magenn Note that the article further says "It may be applied as an operation on propositions, truth values, or semantic values more generally." This makes it suited for quantificational logic as well. $\endgroup$ – user278201 Aug 16 '17 at 20:10
1
$\begingroup$

I think that you made a valid point, but this does not necessarily mean that Rosen made a mistake in applying the negation on $Q(x)$.

What is true is that in quantificational logic, negation is defined in a slightly different way, even though the definition may be carried over in quite an obious way. In propositional logic, we say that if $\alpha$ is a WFF of propositional logic, then $\neg\alpha$ also is one. In quantificational logic, we let $\alpha$ be a WFF of quantificational logic. Since the languages of propositional and quantificational logic differ, so do the definitions of negation.

However, I would assume that Rosen has defined the language of quantifical logic and thereby also redefined negation in a way that makes $\neg Q(x)$ a WFF of quantificational logic.

$\endgroup$
0
$\begingroup$

Negations are defined for statements. Thus applying it on a propositional function is undefined as a propositional function is not a statement. Here it shouldn't have been applied on Q(x).

The predicate, $Q(x)$ (also known as a propositional function), is in fact a statement; that is, a well formed formula.   It can be negated.

The quantified formula, $\exists x{\in}D: Q(x)$, is also a statement.   It too can be negated.   The negation is as supplied.

If it is not so that "Some $x$ in $D$ are such that $Q(x)$ is satisfied" , then it must be so that "Every $x$ in $D$ is such that $Q(x)$ is falsified", and vice versa.

Symbolically, $\neg \big(\exists x {\in} D : Q(x)\big) ~≡~ \big(\forall x {\in} D: \neg Q(x)\big)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.