Propositions like “All humans are mortal” and “Every positive real number has a square-root” are modeled in logic in the form “For all \(x\text{,}\) \(P(x)\)” and “For all \(r\text{,}\) there exists an \(s\) such that \(Q(r,s)\)”, where \(P(x)\) stands for the phrase “\(x\) is mortal” and \(Q(r,s)\) stands for the phrase “\(s\) is a square-root of \(r\)”. Observe that \(P(x)\) and \(Q(r,s)\) on their own are not propositions; there is no truth value to “\(x\) is mortal” or “\(s\) is a square-root of \(r\)”. Instead, we think of \(P(x)\) and \(Q(r,s)\) as functions which return propositions when evaluated at a specific choice for \(x\text{,}\) or for \(r\) and \(s\text{.}\) For example, evaluating \(P(x)\) at \(x=\text{Aaron Greicius}\) yields the proposition “Aaron Greicius is mortal”, which happens to be true at the time of writing. Similarly evaluating \(Q(r,s)\) at \(r=2, s=11\) yields the proposition “\(11\) is a square-root of 2”, which happens to be false. In logic \(P(x)\) and \(Q(r,s)\) are called propositional functions (also called predicates): functions whose outputs are propositions.
Remark 0.2.6. Truth depends on domain of discourse.
Just as a function is not properly defined before its domain is specified, we do not have a well-defined propositional function, nor well-defined truth values of propositions built from this propositional function, until its domain of discourse is given.
For example, let \(P(x)\) be “x>0”. If we declare \(D=(0,\infty)\text{,}\) then the proposition \(\forall x P(x)\) is true, since by definition every \(d\in (0,\infty)\) is positive. On the other hand, if we declare \(D=\R\text{,}\) the proposition \(\forall x P(x)\) is false since not all elements of \(\R\) are positive: indeed, \(-1\) is negative, making \(P(-1)\) false.
Because of the important role played by the domain of discourse \(D\text{,}\) we sometimes include it in our quantifier expressions: e.g., \(\forall x\in D P(x)\text{,}\) \(\exists x\in D P(x)\text{.}\) Using this convention allows us to see more immediately that \(\forall\, x\in (0,\infty)\, P(x)\) is true and \(\forall\, x\in\mathbb{R}\, P(x)\) is false.
The example below taken from calculus nicely illustrates how to negate a proposition that involves a sequence of quantifiers.
Example 0.2.10. The limit does not exist.
Let \(f(x)\) be a function with domain \(\R\text{,}\) and let \(c\in \R\) be a point of this domain. By definition, the proposition that \(\lim\limits_{x\to c}f(x)\) exists is equivalent to the following proposition:
\begin{equation}
\exists L\in\R\, \forall \epsilon \gt 0\, \exists \delta \gt0\, \forall x\in\mathbb{R}\ (\val{x-c}\lt\delta\implies \val{f(x)-L}\lt\epsilon)\text{.}\tag{0.2.1}
\end{equation}
(We made a number of shortcuts in our logical notation above (e.g. \(\forall \epsilon\gt 0\text{,}\) \(\exists\delta\gt 0\)) in order to simplify the expression somewhat; the intended meaning should still be clear. )
Use the negation rules described in
Negating quantifiers to derive a similar proposition equivalent to the statement that
\(\lim\limits_{x\to c}f(x)\) do not exist.
Solution.
Let
\(\mathcal{P}\) be the proposition in
(0.2.1). Using the negation rules in series, we derive the
chain of equivalences below. We’ve added parentheses to emphasize what is being negated at each step. Note how a quantifiers are “swapped” each time we pass the negation to the right.
\begin{align*}
\neg\mathcal{P}\amp\iff \forall L\in\R\, \neg\left(\forall \epsilon \gt 0\, \exists \delta \gt0\, \forall x\in\mathbb{R}\ (\val{x-c}\lt\delta\implies \val{f(x)-L}\lt\epsilon) \right)\\
\amp\iff \forall L\in\R\, \exists \epsilon \gt 0\, \neg\left( \exists \delta \gt0\, \forall x\in\mathbb{R}\ (\val{x-c}\lt\delta\implies \val{f(x)-L}\lt\epsilon) \right)\\
\amp\iff \forall L\in\R\, \exists \epsilon \gt 0\, \forall \delta \gt0\, \neg\left( \forall x\in\mathbb{R}\ (\val{x-c}\lt\delta\implies \val{f(x)-L}\lt\epsilon) \right)\\
\amp\iff \forall L\in\R\, \exists \epsilon \gt 0\, \forall \delta \gt0\, \exists x\in\mathbb{R}\ \neg (\val{x-c}\lt\delta\implies \val{f(x)-L}\lt\epsilon)\\
\amp\iff \forall L\in\R\, \exists \epsilon \gt 0\, \forall \delta \gt0\, \exists x\in\mathbb{R}\ (\val{x-c}\lt\delta \text{ and } \val{f(x)-L}\not\lt\epsilon)\text{.}
\end{align*}
(The last link in our chain uses the fact that \(\neg(\mathcal{Q}\implies\mathcal{R})\) is equivalent to \(\mathcal{Q}\land\neg\mathcal{R}\text{,}\) as a truth table easily shows.) Translating back into English, we conclude that the limit not existing (\(\neg\mathcal{P}\)) is equivalent to the following: for every \(L\in \mathbb{R}\) there is an \(\epsilon\gt 0\) such that for all \(\delta\gt 0\) there exists an \(x\in\mathbb{R}\) satisfying \(\val{x-c}\lt \delta\) and \(\val{f(x)-L}\not\gt\epsilon\text{.}\) Quite a mouthful!