Слияние бета-расширения

Пусть $\to_\beta$ - $\beta$ редукция в $\lambda$ вычислении. Определить $\beta$ расширение $\leftarrow_\beta$ по $t'\leftarrow_\beta t \iff t\to_\beta t'$ .

$\leftarrow_\beta$ является слитым ? Другими словами, имеем ли мы это для любого $l,d,r$ , если $l \to_\beta^* d\leftarrow_\beta^* r$ , то существует такое $u$ , что $l\leftarrow_\beta^* u \to_\beta^* r$ ?

Ключевые слова: восходящее слияние, перевернутая собственность ЧР

Я начал с рассмотрения более слабого свойства: локального слияния (т. Е. Если $l \to_\beta d\leftarrow_\beta r$ , то $l\leftarrow_\beta^* u \to_\beta^* r$ ). Даже если бы это было правдой, это не означало бы$\beta$ расширение не прекращается, но я подумал, что это поможет мне понять препятствия.

(Верх) В случае , когда оба сокращения находятся на верхнем уровне, то гипотеза становится $(\lambda x_1.b_1)a_1\rightarrow b_1[a_1/x_1]=b_2[a_2/x_2]\leftarrow (\lambda x_2.b_2)a_2$ . С точностью до $\alpha$ переименования можно считать, что $x_1\not =x_2$и что ни $x_1$ ни $x_2$ являются свободными в этих терминах.

(Бросок) Если $x_1$ не свободно в $b_1$ , мы имеем $b_1=b_2[a_2/x_2]$ , и , следовательно , имеют $(\lambda x_1.b_1)a_1=(\lambda x_1.b_2[a_2/x_2])a_1\leftarrow(\lambda x_1.(\lambda x_2.b_2)a_2)a_1\rightarrow (\lambda x_2.b_2)a_2$ .

Наивное доказательство по индукции (на $b_1$ и $b_2$ ) для случая (вверху) будет следующим:

• Если $b_1$ является переменной $y_1$ ,

  • Если $y_1=x_1$ , гипотеза становится $(\lambda x_1.x_1)a_1\rightarrow a_1=b_2[a_2/x_2]\leftarrow (\lambda x_2.b_2)a_2$ , и мы действительно имеем $(\lambda x_1.x_1)a_1=(\lambda x_1.x_1)(b_2[a_2/x_2])\leftarrow (\lambda x_1.x_1)((\lambda x_2.b_2)a_2)\rightarrow (\lambda x_2.b_2)a_2$ .

  • Если $y_1\not=x_1$ , то мы можем просто использовать (Throw).

• Те же доказательства применимы, если $b_2$ является переменной.

• Для $b_1=\lambda y.c_1$ и $b_2=\lambda y.c_2$ , гипотеза становится $(\lambda x_1.\lambda y. c_1)a_1\rightarrow \lambda y.c_1[a_1/x_1]=\lambda y.c_2[a_2/x_2]\leftarrow (\lambda x_2.\lambda y.c_2)a_2$ и предположение индукции дает$d$ такимчто$(\lambda x_1.c_1)a_1\leftarrow d\rightarrow (\lambda x_2.c_2)a_2$ откуда следуетчто$\lambda y.(\lambda x_1.c_1)a_1\leftarrow \lambda y.d\rightarrow \lambda y.(\lambda x_2.c_2)a_2$ . К сожалению, у нас нет$\lambda y.(\lambda x_2.c_2)a_2\rightarrow (\lambda x_2.\lambda y.c_2)a_2$ . (Это заставляет меня думать о$\sigma$ сокращении.)

• Аналогичная проблема возникает для приложений: $\lambda$ s не там, где они должны быть.

Два контрпримеры:

• $(\lambda x. b x (b c)) c$ and $(\lambda x. x x) (b c)$ (Plotkin).
• $(\lambda x. a (b x)) (c d)$ and $a ((\lambda y. b (c y)) d)$ (Van Oostrom).

The counterexample detailed below is given in The Lambda Calculus: Its Syntax and Semantics by H.P. Barenredgt, revised edition (1984), exercise 3.5.11 (vii). It is attributed to Plotkin (no precise reference). I give an incomplete proof which is adapted from a proof by Vincent van Oostrom of a different counterexample, in Take Five: an Easy Expansion Exercise (1996) [PDF].

The basis of the proof is the standardization theorem, which allows us to consider only beta expansions of a certain form. Intuitively speaking, a standard reduction is a reduction that makes all of its contractions from left to right. More precisely, a reduction is non-standard iff there is a step $M_i$ whose redex is a residual of a redex to the left of the redex of a previous step $M_j$; “left” and “right” for a redex are defined by the position of the $\lambda$ that is eliminated when the redex is contracted. The standardization theorem states that there if $M \rightarrow_\beta^* N$ then there is a standard reduction from $M$ to $N$.

Let $L = (\lambda x. b x (b c)) c$ and $R = (\lambda x. x x) (b c)$. Both terms beta-reduce to $b c (b c)$ in one step.

Suppose that there is a common ancestor $A$ such that $L \leftarrow_\beta^* A \rightarrow_\beta^* R$. Thanks to the standardization theorem, we can assume that both reductions are standard. Without loss of generality, suppose that $A$ is the first step where these reductions differ. Of these two reductions, let $\sigma$ be the one where the redex of the first step is to the left of the other, and write $A = C_1[(\lambda z. M) N]$ where $C_1$ is the context of this contraction and $(\lambda z. M) N$ is the redex. Let $\tau$ be the other reduction.

Since $\tau$ is standard and its first step is to the right of the hole in $C_1$, it cannot contract at $C_1$ nor to the left of it. Therefore the final term of $\tau$ is of the form $C_2[(\lambda z. M') N']$ where the parts of $C_1$ and $C_2$ to the left of their holes are identical, $M \rightarrow_\beta^* M'$ and $N \rightarrow_\beta^* N'$. Since $\sigma$ starts by reducing at $C_1$ and never reduces further left, its final term must be of the form $C_3[S]$ where the part of $C_3$ to the left of its hole is identical to the left part of $C_1$ and $C_2$, and $M[z \leftarrow N] \rightarrow_\beta^* S$.

Observe that each of $L$ and $R$ contains a single lambda which is to the left of the application operator at the top level. Since $\tau$ preserves the lambda of $\lambda z. M$, this lambda is the one in whichever of $L$ or $R$ is the final term of $\tau$, and in that term the argument of the application is obtained by reducing $N$. The redex is at the toplevel, meaning that $C_1 = C_2 = C_3 = []$.

• If $\tau$ ends in $R$, then $M \rightarrow_\beta^* z z$, $N \rightarrow_\beta^* b c$ and $M[z \leftarrow N] \rightarrow_\beta^* (\lambda x. b x (b c)) c$. If $N$ has a descendant in $L$ then this descendant must also reduce to $b c$ which is the normal form of $N$. In particular, no descendant of $N$ can be a lambda, so $\sigma$ cannot contract a subterm of the form $\check{N} P$ where $\check{N}$ is a descendant of $N$. Since the only subterm of $L$ that reduces to $b c$ is $b c$, the sole possible descendant of $N$ in $L$ is the sole occurrence of $b c$ itself.

• If $\tau$ ends in $L$, then $M \rightarrow_\beta^* b z (b c)$, $N \rightarrow_\beta^* c$, and $M[z \leftarrow N] \rightarrow_\beta^* (\lambda x. x x) (b c)$. If $N$ has a descendant in $R$ then this descendant must also reduce to $c$ by confluence.

At this point, the conclusion should follow easily according to van Oostrom, but I'm missing something: I don't see how tracing the descendants of $N$ gives any information about $M$. Apologies for the incomplete post, I'll think about it overnight.

Note that $\beta$-reduction can make any term disappear. Assuming that variable $x$ does not appear free in a term $v$, you have $(\lambda x.v)\;t_1 \to_\beta v$ and $(\lambda x.v)\;t_2 \to_\beta v$ for any terms $t_1$ and $t_2$. As a consequence, the fact that reverse $\beta$-reduction is confluent is somewhat equivalent to: for all terms $t_1$ and $t_2$, there is a term $u$ such that $u \to_\beta^\ast t_1$ and $u \to_\beta^\ast t_2$. This seems very false to me!