Proof-theoretic description

We let a regularly controlled context-free grammar with appearance checking be a 6-tuple ${\displaystyle G=(N,T,S,P,R,F)}$ where N, T, S, and P are defined as in CFGs, R is a subset of P* constituting a regular language over P, and F is some subset of P. We then define the immediately derives relation ${\displaystyle \Rightarrow _{p_{i}}}$ as follows:

Given some strings x and y, both in ${\displaystyle (N\cup T)^{*}}$, and some rule ${\displaystyle p=A\to w\in P}$,

${\displaystyle x\Rightarrow _{p}^{ac}y}$

holds if either

${\displaystyle x=x_{1}Ax_{2}}$ and ${\displaystyle y=y_{1}wy_{2}}$, or

${\displaystyle x=y}$ and ${\displaystyle p\in F}$

Intuitively, this simply spells out that a rule can apply to a string if the rule's left-hand-side appears in that string, or if the rule is in the set of "vacuously applicable" rules which can "apply" to a string without changing anything. This requirement that the non-vacuously applicable rules must apply is the appearance checking aspect of such a grammar. The language for this kind of grammar is then simply set of terminal strings ${\displaystyle L(G)=\{w\in T^{*}:S\Rightarrow _{p_{1}}^{ac}w_{1}\Rightarrow _{p_{2}}^{ac}w_{2}\Rightarrow _{p_{3}}^{ac}...\Rightarrow _{p_{n}}^{ac}w,\ for\ some\ p_{1}p_{2}...p_{n}\in R\}}$.

Example

Consider a simple (though not the simplest) context-free grammar that generates the language ${\displaystyle \{a^{n}:n\geq 1\}}$:

Let ${\displaystyle G=(\{S,A,X\},\{a\},S,\{f,g,h,k,l\})}$, where

${\displaystyle f:S\to AA}$

${\displaystyle g:S\to X}$

${\displaystyle h:A\to S}$

${\displaystyle k:A\to X}$

${\displaystyle l:S\to a}$

In language controlled form, this grammar is simply ${\displaystyle G^{\prime }=(\{S,A,X\},\{a\},S,\{f,g,h,k,l\},(f|g|h|k|l)^{*},\{f,g,h,k,l\})}$ (where ${\displaystyle (f|g|h|k|l)^{*}}$ is a regular expression denoting the set of all sequences of production rules). A simple modification to this grammar, changing is control sequence set R into the set ${\displaystyle (f^{*}gh^{*}k)^{*}l^{*}}$, and changing its vacuous rule set F to ${\displaystyle \{g,k\}}$, yields a grammar which generates the non-CF language ${\displaystyle \{a^{2^{n}}:n\geq 0\}}$. To see how, consider the general case of some string with n instances of S in it, i.e. ${\displaystyle S^{n}}$ (the special case ${\displaystyle S^{1}}$ trivially derives the string a which is ${\displaystyle a^{2^{0}}}$, an uninteresting fact).

If we chose some arbitrary production sequence ${\displaystyle f^{u}gh^{v}k...}$, we can consider three possibilities: ${\displaystyle n=u}$, ${\displaystyle n<u}$, and ${\displaystyle n>u}$ When ${\displaystyle n=u}$ we rewrite all n instances of S as AA, by applying rule f to the string u times, and proceed to apply g, which applies vacuously (by virtue of being in F) . When ${\displaystyle n<u}$, we rewrite all n instances of S as AA, and then try to perform the n+1 rewrite using rule f, but this fails because there are no more Ss to rewrite, and f is not in F and so cannot apply vacuously, thus when ${\displaystyle n<u}$, the derivation fails. Lastly, then ${\displaystyle n>u}$, we rewrite u instances of S, leaving at least one instance of S to be rewritten by the subsequent application of g, rewriting S as X. Given that no rule of this grammar ever rewrites X, such a derivation is destined to never produce a terminal string. Thus only derivations with ${\displaystyle n=u}$ will ever successfully rewrite the string ${\displaystyle S^{n}}$. Similar reasoning holds of the number of As and v. In general, then, we can say that the only valid derivations have the structure ${\displaystyle S^{n}\Rightarrow _{f}...\Rightarrow _{f}A^{2n}\Rightarrow {g}A^{2n}\Rightarrow {h}...\Rightarrow {h}S^{2n}\Rightarrow {k}S^{2n}}$ will produce terminal strings of the grammar. The X rules, combined with the structure of the control, essentially force all Ss to be rewritten as AAs prior to any As being rewritten as Ss, which again is forced to happen prior to all still later iterations over the S-to-AA cycle. Finally, the Ss are rewritten as as. In this way, the number of Ss doubles each for each instantiation of ${\displaystyle f^{8}gh^{*}k}$ that appears in a terminal-deriving sequence.

Choosing two random non-terminal deriving sequences, and one terminal-deriving one, we can see this in work:

Let ${\displaystyle s_{1}=ffghkll}$, then we get the failed derivation:

${\displaystyle S\Rightarrow _{f}^{ac}AA\Rightarrow _{f}^{ac}{\text{failure: f cannot apply, no S to rewrite}}}$

Let ${\displaystyle s_{2}=fghhhkll}$, then we get the failed derivation:

${\displaystyle S\Rightarrow _{f}^{ac}AA\Rightarrow _{g}^{ac}AA\Rightarrow _{h}^{ac}SA\Rightarrow _{h}^{ac}SS\Rightarrow _{h}^{ac}{\text{failure: h cannot apply, no A to rewrite}}}$

Let ${\displaystyle s_{3}=fghhkll}$, then we get the successful derivation:

${\displaystyle S\Rightarrow _{f}^{ac}AA\Rightarrow _{g}^{ac}AA\Rightarrow _{h}^{ac}SA\Rightarrow _{h}^{ac}SS\Rightarrow _{k}^{ac}SS\Rightarrow _{l}^{ac}aS\Rightarrow _{l}^{ac}aa}$

Similar derivations with a second cycle of ${\displaystyle f^{*}gh^{*}k}$ produce only SSSS. Showing only the (continued) successful derivation:

${\displaystyle ...\Rightarrow SS\Rightarrow _{f}^{ac}AAS\Rightarrow _{f}^{ac}AAAA\Rightarrow _{g}^{ac}AAAA}$

${\displaystyle \Rightarrow _{h}^{ac}SAAA\Rightarrow _{h}^{ac}SSAA\Rightarrow _{h}^{ac}SSSA\Rightarrow _{h}^{ac}SSSS\Rightarrow _{k}^{ac}SSSS}$

${\displaystyle \Rightarrow _{l}^{ac}aSSS\Rightarrow _{l}^{ac}aaSS\Rightarrow _{l}^{ac}aaaS\Rightarrow _{l}^{ac}aaaa}$

Example

As with so many controlled grammars, programmed grammars can generate the language ${\displaystyle \{a^{2^{n}}:n\geq 0\}}$:

Let ${\displaystyle G=(\{S,A\},\{a\},S,\{r_{1},r_{2},r_{3}\})}$, where

${\displaystyle r_{1}=(S\to AA,\{r_{1}\},\{r_{2}\})}$

${\displaystyle r_{2}=(A\to S,\{r_{2}\},\{r_{1},r_{3}\})}$

${\displaystyle r_{3}=(S\to a,\{r_{3}\},\emptyset )}$

The derivation for the string aaaa is as follows:

${\displaystyle S\Rightarrow _{r_{1}}AA\Rightarrow _{r_{1}}AA\Rightarrow _{r_{2}}SA\Rightarrow _{r_{2}}SS\Rightarrow _{r_{2}}SS}$

${\displaystyle \Rightarrow _{r_{1}}AAS\Rightarrow _{r_{1}}AAAA\Rightarrow _{r_{1}}AAAA}$

${\displaystyle \Rightarrow _{r_{2}}SAAA\Rightarrow _{r_{2}}SSAA\Rightarrow _{r_{2}}SSSA\Rightarrow _{r_{2}}SSSS\Rightarrow _{r_{2}}SSSS}$

${\displaystyle \Rightarrow _{r_{3}}aSSS\Rightarrow _{r_{3}}aaSS\Rightarrow _{r_{3}}aaaS\Rightarrow _{r_{3}}aaaa\Rightarrow _{r_{3}}aaaa}$

As can be seen from the derivation and the rules, each time ${\displaystyle r_{1}}$ and ${\displaystyle r_{2}}$ succeed, they feed back to themselves, which forces each rule to continue to rewrite the string over and over until it can do so no more. Upon failing, the derivation can switch to a different rule. In the case of ${\displaystyle r_{1}}$, that means rewriting all Ss as AAs, then switching to ${\displaystyle r_{2}}$. In the case of ${\displaystyle r_{2}}$, it means rewriting all As as Ss, then switching either to ${\displaystyle r_{1}}$, which will lead to doubling the number of Ss produced, or to ${\displaystyle r_{3}}$ which converts the Ss to as then halts the derivation. Each cycle through ${\displaystyle r_{1}}$ then ${\displaystyle r_{2}}$ therefore either doubles the initial number of Ss, or converts the Ss to as. The trivial case of generating a, in case it is difficult to see, simply involves vacuously applying ${\displaystyle r_{1}}$, thus jumping straight to ${\displaystyle r_{2}}$ which also vacuously applies, then jumping to ${\displaystyle r_{3}}$ which produces a.

Proof-theoretic definition

With this definition of a conditional grammar, we can define the derives relation as follows:

Given two strings ${\displaystyle x,y\in (N\cup T)^{*}}$, and some production rule ${\displaystyle p=(A\to w,R)\in P}$,

${\displaystyle x\Rightarrow _{p}y}$ if and only if ${\displaystyle x=x'Ax''}$, ${\displaystyle y=x'wx''}$, and ${\displaystyle x\in R}$

Informally then, the production rule for some pair in P can apply only to strings that are in its context language. Thus, for example, if we had some pair ${\displaystyle (S\to x,a^{*}Sb^{*})}$, we can only apply this to strings consisting of any number of as followed by exactly only S followed by any number of bs, i.e. to sentences in ${\displaystyle \{a^{m}Ab^{n}:m,n\geq 0\}}$, such as the strings S, aSb, aaaS, aSbbbbbb, etc. It cannot apply to strings like xSy, aaaSxbbb, etc.

Example

Conditional grammars can generate the context-sensitive language ${\displaystyle \{a^{2^{n}}:n\geq 0\}}$.

Let ${\displaystyle G=(\{S,S'\},\{a\},\{f,g,h\},S)}$, where

${\displaystyle f=(S\to AA,A^{*}S^{+})}$

${\displaystyle g=(A\to B,B^{*}A^{+})}$

${\displaystyle h=(B\to S,S^{*}B^{+})}$

${\displaystyle k=(S\to a,a^{*}S^{+})}$

We can then generate the sentence aaaa with the following derivation:

${\displaystyle S\Rightarrow _{f}AA\Rightarrow _{g}BA\Rightarrow _{g}BB}$

${\displaystyle \Rightarrow _{h}SB\Rightarrow _{h}SS\Rightarrow _{f}AAS\Rightarrow _{f}AAAA}$

${\displaystyle \Rightarrow _{g}BAAA\Rightarrow _{g}BBAA\Rightarrow _{g}BBBA\Rightarrow _{g}BBBB}$

${\displaystyle \Rightarrow _{h}SBBB\Rightarrow _{h}SSBB\Rightarrow _{h}SSSB\Rightarrow _{h}SSSS}$

${\displaystyle \Rightarrow _{k}aSSS\Rightarrow _{k}aaSS\Rightarrow _{k}aaaS\Rightarrow _{k}aaaa}$

Example

Like conditional grammars, random context grammars (and thus semi-conditional grammars) can generate the language ${\displaystyle \{a^{2^{n}}:n\geq 0\}}$. One grammar which can do this is:

Let ${\displaystyle G=(\{S,X,Y,A\},\{a\},S,\{r_{1},r_{2},r_{3},r_{4},r_{5}\})}$, where

${\displaystyle r_{1}=(S\to XX,\emptyset ,\{Y,A\})}$

${\displaystyle r_{2}=(X\to Y,\emptyset ,\{S\})}$

${\displaystyle r_{3}=(Y\to S,\emptyset ,\{X\})}$

${\displaystyle r_{4}=(S\to A,\emptyset ,\{X,Y\})}$

${\displaystyle r_{5}=(A\to a,\emptyset ,\{S\})}$

Consider now the production for aaaa:

${\displaystyle S\Rightarrow _{r_{1}}XX\Rightarrow _{r_{2}}YX\Rightarrow _{r_{2}}YY\Rightarrow _{r_{3}}SY\Rightarrow _{r_{3}}SS}$

${\displaystyle \Rightarrow _{r_{1}}XXS\Rightarrow _{r_{1}}XXXX\Rightarrow _{r_{2}}YXXX\Rightarrow _{r_{2}}YYXX\Rightarrow _{r_{2}}YYYX\Rightarrow _{r_{2}}YYYY}$

${\displaystyle \Rightarrow _{r_{3}}SYYY\Rightarrow _{r_{3}}SSYY\Rightarrow _{r_{3}}SSSY\Rightarrow _{r_{3}}SSSS}$

${\displaystyle \Rightarrow _{r_{4}}ASSS\Rightarrow _{r_{4}}AASS\Rightarrow _{r_{4}}AAAS\Rightarrow _{r_{4}}AAAA}$

${\displaystyle \Rightarrow _{r_{5}}aAAA\Rightarrow _{r_{5}}aaAA\Rightarrow _{r_{5}}aaaA\Rightarrow _{r_{5}}aaaa}$

The behavior of the R sets here is trivial: any string can be rewritten according to them, because they do not require any substrings to be present. The behavior of the Q sets, however, are more interesting. In ${\displaystyle r_{1}}$, we are forced by the Q set to rewrite an S, thus beginning an S-doubling process, only when no Ys or As are present in the string, which means only when a prior S-doubling process has been fully initiated, eliminating the possibility of only doubling some of the Ss. In ${\displaystyle r_{2}}$, which moves the S-doubling process into its second stage, we cannot begin this process until the first stage is complete and there are no more Ss to try to double, because the Q set prevents the rule from applying if there is an S symbol still in the string. In ${\displaystyle r_{3}}$, we complete the doubling stage by introducing the Ss back only when there are no more Xs to rewrite, thus when the second stage is complete. We can cycle through these stages as many times as we want, rewriting all Ss to XXs before then rewriting each X to a Y, and then each Y to an S, finally ending by replacing each S with an A and then an a. Because the rule for replacing S with A prohibits application to a string with an X in it, we cannot apply this in the middle of the first stage of the S-doubling process, thus again preventing us from only doubling some Ss.

Example

Like many other contextually controlled grammars, ordered grammars can enforce the application of rules in a particular order. Since this is the essential property of previous grammars that could generate the language ${\displaystyle \{a^{2^{n}}:n\geq 0\}}$, it should be no surprise that a grammar that explicitly uses rule ordering, rather than encoding it via string contexts, should similarly be able to capture that language. And as it turns out, just such an ordered grammar exists:

Let ${\displaystyle G=(\{S,X,Y,Z,A\},\{a\},S,P)}$, where P is the partially ordered set described by the Hasse diagram

The derivation for the string aaaa is simply:

${\displaystyle S\Rightarrow _{S\to XX}\ XX\ \Rightarrow _{X\to Y}\ YX\ \Rightarrow _{X\to Y}\ YY\ \Rightarrow _{Y\to S}\ SY\ \Rightarrow _{Y\to S}\ YY}$

${\displaystyle \Rightarrow _{S\to XX}\ XXS\ \Rightarrow _{S\to XX}\ XXXX}$

${\displaystyle \Rightarrow _{X\to Y}\ YXXX\ \Rightarrow _{X\to Y}\ YYXX\ \Rightarrow _{X\to Y}\ YYYX\ \Rightarrow _{X\to Y}\ YYYY}$

${\displaystyle \Rightarrow _{Y\to S}\ SYYY\ \Rightarrow _{Y\to S}\ SSYY\ \Rightarrow _{Y\to S}\ SSSY\ \Rightarrow _{Y\to S}\ SSSS}$

${\displaystyle \Rightarrow _{S\to A}\ ASSS\ \Rightarrow _{S\to A}\ AASS\ \Rightarrow _{S\to A}\ AAAS\ \Rightarrow _{S\to A}\ AAAA}$

${\displaystyle \Rightarrow _{A\to a}\ aAAA\ \Rightarrow _{A\to a}\ aaAA\ \Rightarrow _{A\to a}\ aaaA\ \Rightarrow _{A\to a}\ aaaa}$

At each step of the way, the derivation proceeds by rewriting in cycles. Notice that if at the fifth step SY, we had four options: ${\displaystyle Y\to Z,S\to Z,Y\to S,S\to A}$, the first two of which halt the derivation, as Z cannot be rewritten. In the example, we used ${\displaystyle Y\to S}$ to derive SS, but consider if we had chosen ${\displaystyle S\to A}$ instead. We would have produced the string AS, the options for which are ${\displaystyle Y\to Z}$ and ${\displaystyle A\to Z}$, both of which halt the derivation. Thus with the string SY, and conversely with YS, we must rewrite the Y to produce SS. The same hold for other combinations, so that overall, the ordering forces the derivation to halt, or else proceed by rewriting all Ss to XXs, then all Xs to Ys, then all Ys to Ss, and so on, then finally all Ss to As then all As to as. In this way, a string ${\displaystyle S^{n}}$ can only ever be rewritten as ${\displaystyle A^{n}}$ which produces as, or as ${\displaystyle S^{2n}}$. Starting with n = 0, it should be clear that this grammar only generates the language ${\displaystyle \{a^{2^{n}}:n\geq 0\}}$.

Example

Scattered context grammars are capable of describing the language ${\displaystyle \{a^{n}b^{n}c^{n}:n\geq 0\}}$ quite easily.

Let ${\displaystyle G=(\{S,A,B,C\},\{a,b,c\},S,\{r_{1},r_{2},r_{3}\})}$, where

${\displaystyle r_{1}=(S\to ABC)}$

${\displaystyle r_{2}=(A\to aA,B\to bB,C\to cC)}$

${\displaystyle r_{3}=(A\to \epsilon ,B\to \epsilon ,C\to \epsilon )}$

Deriving aaabbbccc then is trivial:

${\displaystyle S\Rightarrow _{r_{1}}ABC\Rightarrow _{r_{2}}aAbBcC\Rightarrow _{r_{2}}aaAbbBccC\Rightarrow _{r_{2}}aaaAbbbBcccC\Rightarrow _{r_{3}}aaabbbccc}$