Does "if" turn a free variable into a bound variable?

Question

A closed formula expresses a proposition (which is a truth-apt concept, so a concept that is either be "true" or "false"). A closed formula is a Boolean-valued formula with no free variables.

A free variable is a variable that "specifies places in an expression where substitution may take place and is not a parameter of this or any container expression", it has not been bound by a variable-binding operator like

$$ \sum _{x\in S}\quad \quad \prod _{x\in S}\quad \quad \int _{0}^{\infty }\cdots \,dx\quad \quad \lim _{x\to 0}\quad \quad \forall x\quad \quad \exists x $$

So for example (in the following all numbers, variables, and function values are real numbers)

\begin{align} f(x) = 5, f(x) = 2*x \qquad\qquad\qquad\qquad\qquad (1) \end{align}

is not true or false, as long as we don't know what $x$ is. But if we say,

$$ \text{if } x = 2, f(x) = 2*x \text{, then } f(x) = 5 \qquad\;\;\qquad (2) $$

this becomes truth-apt (in this case this is false). So, I would think that "if" is a variable-binding operator. Is this correct?

Another way of writing "if" is by using $\Rightarrow$, e.g.,

$$ x = 2, f(x) = 2*x \Rightarrow f(x) = 5 \qquad\qquad\qquad (3) $$

which I would think is a closed formula which is false.

Answer 1

\begin{align} f(x) = 5\text{ and } f(x) = 2x\tag1 \end{align} is not true or false, as long as we don't know what $x$ is.

$$\text{if } x = 2\text{ and } f(x) = 2x \text{, then } f(x) = 5\tag2$$ this becomes truth-apt (in this case this is false). So, I would think that "if" is a variable-binding operator. Is this correct?

Has $f$ been pre-defined such that $f(x)=2x\,?$ If so, then the open formula $(2)$ is actually true for all values of $x$ except $2.5.$ What is false though is the closed formula $$\text{for each $x,\Big($if } x = 2\text{ and } f(x) = 2x \text{, then } f(x) = 5\Big)\tag{2a}.$$ Informally, open conditional formulae like $(2)$ are typically treated as closed formulae like $(2a);$ it is this implicit universal quantification—not the conditional operator per se—that binds the free variable $x.$

On the other hand, if $f(x)$ actually equals, say, $-\dfrac1x,$ then both formulae $(2)$ and $(2a)$ are definitely true in the standard interpretation.

Answer 2

Actually, we do not need variable-bounding in this instance; quantifier-free fragment of first-order logic with equality (see Ian Chiswell and Wilfrid Hodges' Mathematical Logic, chapter 5) or equational logic will do.

We can take the signature as

$\sigma_{eq} = \langle \mathcal{A}, a, b, c,\ldots f, g, h,\ldots =, R, S,\ldots\rangle$

where $\mathcal{A}$ is a set as the domain of discourse; $a, b, c,\ldots$ are individual constants; $f, g, h,\ldots$ are (term-forming) functions; $=, R, S,\ldots$ are relations (representing predicates in the domain). '$=$' is the equality relation, it can be put into the logical vocabulary into the signature.

The functions can be represented as

$\phi$ as $=\!(x, 2)$, $\psi$ as $=\!(f(x), 2)$, $\upsilon$ as $=\!(f(x), 5)$.

Natural deduction system with the '=' introduction rule

$\dfrac{}{t=t}\;\;\; (=I)$

and '=' elimination rule

$\dfrac{s = t\;\;\;\;\phi(s/x)}{\phi(t/x)}\;\;\; (=E)$

together with the equality axioms of reflexivity, symmetry, and transitivity, will suffice.

A significant point in this perspective is that free variables are not merely syntactic elements. They are also semantic constituents that serve roughly as names or parameters in a fashion comparable to logical names in computer science for which we have not specified definite physical names.