So I am having some confusion about the notation that is used in "Introduction to Mathmatical Logic" by Elliot Mendelson for indicating that some of the variables are free in a wf $\mathscr B$.[Below exercise 2.5]

We shall often indicate that some of the variables $x_{i_1},...,x_{i_k}$ are free variables in a wf $\mathscr B$ by writing $\mathscr B$ as $\mathscr B(x_{i_1},...,x_{i_k})$ .This does not mean that $\mathscr B$ contains these variables as free variables , nor does it mean that $\mathscr B$ does not contain other free variables.

In order for a variable to be a free variable in $\mathscr B$ , it has to have an free occurrence in the wf $\mathscr B$ atleast once.So if a variable $x$ doesn't exist in a wf $\mathscr B$ , then it is not a free variable in $\mathscr B$.
But , the above passage allows us to indicate these "non existent variables" as free variables in $\mathscr B$ by using the notation $\mathscr B(x_{i_1},...,x_{i_k})$. Is this right to do?

There is another confusion of mine regarding non-Existent variables. Lets say $y$ is a variable that doesn't exist in $\mathscr B$ .Then because $\mathscr B$ doesn't have $y$ as a variable , so we can write $\mathscr B \equiv \forall y\mathscr B$ according to this passage.

Notice that $\mathscr B$ need not contain the variable $y$. In that case, we understand (($\forall y)\mathscr B$) to mean the same thing as $\mathscr B$

In this case , $y$ is bound in $\forall y \mathscr B$ .And because $\mathscr B \equiv \forall y\mathscr B$ , we can say that $y$ is bound in $\mathscr B$.Is this faulty?

  • $\begingroup$ It is only a convention. See also van Dalen $\endgroup$ – Mauro ALLEGRANZA Feb 23 at 7:35
  • $\begingroup$ If it is disturbing for you, you can imagine $B':=B\land(x_1=x_1\land x_2=x_2\land\dots)$. $\endgroup$ – Berci Feb 23 at 8:18
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    $\begingroup$ Again, the issue is the difference between a formula of the language: $A_1^2(x_1,x_2)$ and a schematic expression of the meta-language. Having said that, see page 49: "An occurrence of a variable $x$ is said to be bound in a wf $\mathscr B$ if either it is the occurrence of $x$ in a quantifier “$(∀x)$” in $\mathscr B$ or it lies within the scope of a quantifier “$(∀x)$” in $\mathscr B$. Otherwise, the occurrence is said to be free in $\mathscr B$." Thus, if var $x_1$ does not occur in the formula it is neither free nor bound. $\endgroup$ – Mauro ALLEGRANZA Feb 23 at 10:13
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    $\begingroup$ Mendelson's convention is rather loose but suggestive: $\mathcal{B}(x_{i_1}, \ldots, x_{i_k})$ is just an indicator that some of the $x_{i_j}$ may appear free in $\mathcal{B}$. This allows him to write $\mathcal{B}(t_1, \ldots, t_k)$ for the result of substituting a list of terms $t_j$ for the $x_{i_j}$, which makes it reasonably clear what is substituted for what, but doesn't imply that all the $x_{ij}$ actually appear free. (So he can describe substitution of $1$ for $x$ and $2$ for $y$ in $x = x \land z = 3$, even though $y$ doesn't appear free in the formula, while $z$ does.) $\endgroup$ – Rob Arthan Feb 23 at 11:31
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    $\begingroup$ If $y$ is not free in $\mathcal{B}$, then $\forall y \mathcal{B}$ and $\mathcal{B}$ have the same meaning (they are logically equivalent) but they are not syntactically identical. You can write $\mathcal{B} \equiv \forall y \mathcal{B}$ (if you are using $\equiv$ for logical equivalence, which I don't think Mendelson does), but it doesn't mean that the two formulas have the same syntactic properties and whether or not a variables occurs bound in a formula is a syntactic property. $\endgroup$ – Rob Arthan Feb 23 at 12:38

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