### Example 1

Let $s_1$ be the string “$\forall x:x+2=5$”, let $s_2$ be the string “$\exists x:x+2=5$”, let $U$ be the domain of discourse of natural numbers $0,1,2,\ldots,$ and let $I(s)$ be the interpretation of $s$ assigning it a meaning of adding natural numbers in $U$. Then the strings mean the following:

$s_1$: “For all natural numbers $x=0,1,2,\ldots$ the equation $x+2=5$ holds.”

$s_2$: “There exists a natural number $x$ such that the equation $x+2=5$ holds.”

Clearly, only $s_2$ is true. Thus the valuation function gives us $[[s_1]]_I=0$ and $[[s_2]]_I=1.$

### Example 2

For the same domain of discourse and the same interpretation as in example 1, let $s$ be the string “$\exists x: x+2=y$.” In this case, $x$ is a bound variable and $y$ is a free variable. The string means the following:

$s$: “There exists a natural number $x$ such that the equation $x+2=y$ holds.”

The valuation function $[[s]]_I=undefined$, since the free variable $y$ prevents us from deciding if $s$ is true or false. We won’t be able to decide, unless we know, which natural number is assigned to the free variable $y.$

### Example 3

Take real numbers as the domain of discourse, and consider the \(\epsilon-\delta\) definition of continuous real functions:

A real function \(f:D\to\mathbb R\) is continuous at the point \(a\in D\), if for every \(\epsilon > 0\) there is a \(\delta > 0\) such that \(|f(x)-f(a)| < \epsilon\) for all \(x\in D\) with \(|x-a| < \delta.\)

This proposition can be codified using a string like this:

“$\forall\,(\epsilon > 0)\,\exists\,(\delta > 0)\,\forall\,(x\in D)\,(|x-a| < \delta\Longrightarrow|f(x)-f(a)| < \epsilon).$”

In this string, the substrings $\epsilon$, $\delta$ and $x$ are bound variables, since they occur together with the quantifiers $\exists$ and $\forall$, whereas $a$ is a free variable since it never occurs with a quantifier.

| | | | created: 2018-02-11 14:29:03 | modified: 2018-02-11 15:27:09 | by: *bookofproofs*