Undefined term while negating universal or existential statements.

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.