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On page 7/8, section 1.2, of Practical Foundations of Programming Languages, 2nd edition, Robert Harper gives this initial definition of abstract binding trees:

The smallest family of sets closed under the conditions

  1. If $x \in \mathcal{X}_s$, then $x \in \mathcal{B}[\mathcal{X}]_s$
  2. For each operator $o$ of arity $(\vec{s_1}.s_1,\ldots,\vec{s_n}.s_n)s$, if $a_1 \in \mathcal{B}[\mathcal{X},\vec{x_1}]_{s_1},\,\ldots,\, a_n \in \mathcal{B}[\mathcal{X},\vec{x_n}]_{s_n}$, then $o(\vec{x_1}.a_1;\ldots;\vec{x_n}.a_n) \in \mathcal{B}[\mathcal{X}]$

(Here $\mathcal{X}$ denotes a set of variables, $\mathcal{X},x$ the union of $\mathcal{X}$ with $\{x\}$ where $x$ is fresh for $\mathcal{X}$, $\vec{x}$ a sequence of variables,$\mathcal{X}_s$ a set of variables of sort $s$, $\mathcal{B}[X]_s$ the set of abstract binding trees of sort $s$ over the variables in $\mathcal{X}$

This definition is almost correct, but fails to properly account for renaming of bound variables. An abt of the form $\text{let}(a_1;x.\text{let}(a_2;x.a_3))$ is ill-formed according to this defnition, because the first binding adds $x$ to $\mathcal{X}$, which implies that the second cannot also add $x$ to $\mathcal{X},x$, because it is not fresh for $\mathcal{X},x$.

I am confused about his meaning here.

How does this definition result in an ill-formed abt?

By first/second binding does he mean A) outer/inner (read from left to right) or B) inner/outer (read from the inside out)?

What I think he is saying:
Because of the outer("first") binding of $x$, assume that $x$ occurs free in $a_2$. For example $a_2=x,\, a_3=x$. Then because $x$ occurs free in $a_2$, it must be that $a_2 \in \mathcal{B}[\mathcal{X}]$ where $\mathcal{X} = \{x\}$. Since $a_3$ occurs inside an abstractor that binds $x$, $a_3 \in \mathcal{B}[\mathcal{X,x}]$, but then $a_3 \in \mathcal{B}[\{x\},x]$ which is ill-formed since $x$ is not fresh for $\{x\}$

But then I think of the concrete example $\text{let}(y,x.\text{let}(z,x.x))$ in which $a_2 \in \mathcal{B}[\{z\}]$ and $a_3 \in \mathcal{B}[\{z\},x]$, which poses no problems in this interpretation.


Edit to elaborate on the accepted answer...

What I now believe Harper meant is that the outer binding of $x$ indicates that $x$ is considered to be among the free, or "already used" variables in the inner let. This may or may not mean that $x$ must actually appear free in the inner let.

In either case, it means that validation of abts for well-formedness must proceed from the outside-in. In the specific examples Harper gives, the outer binding of $x$ means $x \in X$ in the validation of the inner let:
if $a_2 \in \mathcal{B}[\mathcal{X}]$ and $a_3 \in \mathcal{B}[\mathcal{X},x] \ldots$ (<-- ill formed; $x$ is not fresh for $\mathcal{X}$)

If in particular the wording means that $x$ must specifically appear free in the inner let, then in the given example, it would have to be in $a_2$ as suggested in my question and in the answer below. This amounts to saying that a particular instance of an abt in $\mathcal{X}$ is not automatically an abt in $\mathcal{X}\cup\mathcal{Y}$ for any set of variables $\mathcal{Y}$.

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afsmi
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2 Answers2

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Just to clear something up that may not have been obvious, $\chi$ is a set and the notation $B[\chi, x]$ is meant to be ABTs under free variables that are either $x$ or are in $\chi$. In this notation I believe it is implied that $x \notin \chi$ when you write $B[\chi, x]$, which is important.

Using the definition of ABT above, you cannot prove for any $\chi$ that let(z, x.x) is in $B[\chi,x]$, but i claim that this is necessary to prove if you want to use it as the inner formula $a_1$ in let(y, x.$a_1$).

The reason you cannot prove let(z, x.x) is in $B[\chi,x]$ is because using the rule#1 stated above, the free variable $x$ is only an ABT in $B[\chi]$ when $x \in \chi$ (or alternatively: $x$ is only an ABT in $B[\chi',x]$ for some $\chi'$), but then using rule#2, we deduce $z \in B[\chi] \wedge x \in B[\chi, x] \rightarrow let[z,x.x] \in B[\chi]$. As I had mentioned in the first paragraph, this implies $x \notin \chi$, which means that the formula let(z, x.x) is ONLY in ABTs where $x$ is not a free variable.

The reason why $x$ needs to be free in this inner formula is because otherwise we cant apply rule#2 again on the outer formula let(y, x.$a_1$).

Kurt Mueller
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Another attempt at explaining why the first definition doesn't work, so assume that we are using that definition.

Let the only sort be $\mathsf{Exp}$, let $\mathcal{X}_{\mathsf{Exp}} = \{y,z\}$, and let $x$ be a variable (of sort $\mathsf{Exp}$) that is fresh for $\mathcal{X}$. Then $y,z \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}$ and $x \in \mathcal{B}[\mathcal{X},x]_{\mathsf{Exp}}$, which implies that

$$ \mathtt{let}(y; x.\!x) \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}, $$

since $\mathtt{let}$ has (generalised) arity $(\mathsf{Exp}, \mathsf{Exp}.\!\mathsf{Exp})\mathsf{Exp}$. In order to show (the false proposition) that

$$ \mathtt{let}(z; x.\!\mathtt{let}(y; x.\!x)) \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}, $$

we would need $\mathtt{let}(y; x.\!x)$ to not only be in $\mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}$, but also in $\mathcal{B}[\mathcal{X,x}]_{\mathsf{Exp}}$. And this is what is impossible, since this requires $x$ to be fresh for $\mathcal{X},x$. (For ABTs with fresh renaming $\mathcal{X} \subseteq \mathcal{Y}$ implies $\mathcal{B}[\mathcal{X}] \subseteq \mathcal{B}[\mathcal{Y}]$, cf. Exercise 1.2, but this doesn't hold for ABTs without fresh renaming.)

Instead use the second definition of ABTs. Let $\rho(x) = x'$ be a fresh renaming of $x$ (relative to $\mathcal{X}$), and let $\rho'(x') = x''$ be a fresh renaming relative to $\mathcal{X},x$. Then $\hat{\rho}'(x') = \rho(x') = x'' \in \mathcal{B}[\mathcal{X},x',x'']_{\mathsf{Exp}}$, so we have

$$ \mathtt{let}(y; x'\!.\!x') \in \mathcal{B}[\mathcal{X},x']_{\mathsf{Exp}}, $$

since $\rho'$ was an arbitrary fresh renaming of $x'$. Next we have

$$ \hat{\rho}(\mathtt{let}(y; x.\!x)) = \mathtt{let}(y; x'\!.\!x') \in \mathcal{B}[\mathcal{X},x']_{\mathsf{Exp}}, $$

and since $\rho$ was an arbitrary fresh renaming of $x$, this implies that

$$ \mathtt{let}(z; x.\!\mathtt{let}(y; x.\!x)) \in \mathcal{B}[\mathcal{X}]_{\mathsf{Exp}}. $$

Danny
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