Yu. I. Yanov

ON IDENTICAL TRANSFORMATIONS OF REGULAR EXPRESSIONS

(Presented by Academician P. S. Novikov on 6 VI 1962)

Let there be an alphabet \(A_s=\{a_1,\ldots,a_s\}\), \(1 \leq s \leq \aleph_0\). We shall consider sets of (finite) words in this alphabet. The sets \(\{\lambda\}\) and \(\{a_i\}\), \(i=1,\ldots,s\), where \(\lambda\) is the empty word, will be called elementary. Consider the following three basic operations on sets of words:

1) union \(A \vee B\)—the set-theoretic sum of \(A\) and \(B\);

2) product \(AB\)—the set of all words of the form \(ab\), where \(a \in A\), \(b \in B\);

3) closure
\[ \overline{A}=\{\lambda\}\vee A\vee AA\vee AAA\vee\cdots \]

Sets that can be obtained from elementary ones by a finite number of applications of the basic operations will be called regular sets. The class of all regular sets for the alphabet \(A_s\) will be denoted by \(\mathfrak{M}_s\). It is known \((^1)\) that regular sets, and only they, are the sets representable by finite automata.

We shall omit braces in the notation of one-element sets; then every regular set will be given by some formula representing a superposition of the basic operations, whose arguments are the letters of the original alphabet \(A_s\) and, possibly, the symbol of the empty word \(\lambda\). Such formulas will be called regular expressions in the alphabet \(A_s\).

It is obvious that one and the same regular set can be represented by different regular expressions; thus, for example, the regular expressions
\[ \overline{(a_1a_2\vee a_1a_2)(a_3a_4a_5\vee a_3a_6)} \quad\text{and}\quad \overline{a_1a_2\vee a_3a_6\vee a_4\vee a_5} \]
define one and the same set. In this connection there arises the problem of identical transformations of regular expressions in the alphabet \(A_s\). It can be shown that this problem reduces to finding a complete system of identities for the algebra
\[ \mathfrak{R}_s=\langle \mathfrak{M}_s,\ x\vee y,\ xy,\ \overline{x}\rangle.^* \]
If the empty word is disregarded, then the above problem reduces to finding a complete system of identities for the algebra
\[ \mathfrak{R}'_s=\langle \mathfrak{M}'_s,\ x\vee y,\ xy,\ \overline{x}\rangle, \]
where \(\mathfrak{M}'_s\) is the class of all those and only those regular sets (in the alphabet \(A_s\)) which contain the empty word.

In the present work finite complete systems of identities are constructed for the algebras \(\mathfrak{R}'_s\), \(1 \leq s \leq \aleph_0\). It turns out that for any \(p\) and \(q\) such that \(2 \leq p,q \leq \aleph_0\), the algebras \(\mathfrak{R}'_p\) and \(\mathfrak{R}'_q\) are similar, i.e. have identical systems of identities.

Consider the following system of identities \(\Sigma'\):

\[ \begin{array}{ll} 0.\quad x=x. & 8.\quad \overline{x}x=\overline{x}.\\[2mm] 1.\quad x\vee x=x. & 9.\quad \overline{x\vee y}\vee x=\overline{x\vee y}.\\[2mm] 2.\quad x\vee y=y\vee x. & 10.\quad \overline{x\vee y}=\overline{\overline{x}\vee y}.\\[2mm] 3.\quad x\vee (y\vee z)=(x\vee y)\vee z. & 11.\quad \overline{x\vee y}=\overline{xy}.\\[2mm] 4.\quad x(yz)=(xy)z. & 12.\quad \overline{xyx}=\overline{xy}.\\[2mm] 5.\quad x(y\vee z)=xy\vee xz. & 13.\quad x\overline{xy}=\overline{xy}.\\[2mm] 6.\quad (x\vee y)z=xz\vee yz. & 14.\quad xy\vee y=xy.\\[2mm] 7.\quad \overline{\overline{x}}=\overline{x}. & 15.\quad xy\vee x=xy. \end{array} \]

* As V. N. Red’ko has shown, the algebras \(\mathfrak{R}_s\) do not have finite complete systems of identities.

It is not hard to convince oneself that all identities of this system are true for any algebra \(\mathfrak R'_s\), \(1 \leq s \leq \aleph_0\). We shall show that for every algebra \(\mathfrak R'_s\), where \(2 \leq s \leq \aleph_0\), the system \(\Sigma'\) is complete, i.e., every identity true in \(\mathfrak R'_s\) is derivable from \(\Sigma'\) by means of the following two rules of inference:

\[ \alpha^0.\quad \frac{R_1(x_1,\ldots,x_i,\ldots,x_m)=R_2(x_1,\ldots,x_i,\ldots,x_m)} {R_1(x_1,\ldots,S,\ldots,x_m)=R_2(x_1,\ldots,S,\ldots,x_m)} \qquad \text{(substitution rule),} \]

\[ \alpha^1.\quad \frac{R_1(S_1)=R_2,\; S_1=S_2} {R_1(S_2)=R_2} \qquad \text{(replacement rule).} \]

A complete system of identities for the algebra \(\mathfrak R'_1\) can be obtained by adding to \(\Sigma'\) the following two identities:

\[ xy=yx,\qquad \overline{xy}=\bar x\bar y. \]

We shall denote by the symbol \(\vdash\) derivability by means of the rules \(\alpha^0,\alpha^1\). We note that from \(\Sigma'\) the following identities and rules are derivable:

\[ 9'.\quad \overline{xy}\vee x=\overline{xy}. \qquad 9''.\quad \overline{xy}\vee \bar x=\overline{xy}. \]

\[ \gamma.\quad \frac{R_1\vee S_1=S_1,\; R_2\vee S_2=S_2} {R_1R_2\vee S_1S_2=S_1S_2}. \qquad \varepsilon.\quad \frac{R\vee S=S} {\bar R\vee S\vee P=S\vee P}. \]

\[ \delta.\quad \frac{\bar R\vee S=S} {R\vee S=S}. \qquad \zeta.\quad \frac{R_1\vee S=S,\; R_2\vee S=S} {R_1\vee R_2\vee S=S}. \]

All subsequent arguments are valid for any algebra \(\mathfrak R'_s\), where \(2 \leq s \leq \aleph_0\).

Let \(R\) and \(S\) be formulas of the algebra \(\mathfrak R'_s\). We shall write: \(R \subseteq S\), if \(R\vee S=S\) is an identity true for \(\mathfrak R'_s\).

Obviously, in order to prove the completeness of the system \(\Sigma'\), it is sufficient for us to prove the following assertion:

У1. For any formulas \(R\) and \(S\), if \(R\subseteq S\), then \(\Sigma'\vdash R\vee S=S\).

Formulas of the form

\[ \overline{x_{i_1}x_{i_2}\ldots x_{i_m}}, \]

where \(i_1<i_2<\cdots<i_m\), will be called cycles. Variables and cycles will be called elementary chains. Products of elementary chains will be called chains. The empty chain, as well as chains containing no cycles, will be called simple chains. Simple chains will also be regarded by us as words in the alphabet of variables.

For each formula \(R\) we define the set \(\mathfrak S(R)\) of simple chains, which we shall call \(R\)-words.

1) \(\mathfrak S(x_i)=\{\lambda,x_i\}\).

2) Suppose \(\mathfrak S(R_1)\) and \(\mathfrak S(R_2)\) have been defined; then:

2,1) \(\mathfrak S(R_1\vee R_2)=\mathfrak S(R_1)\vee \mathfrak S(R_2)\);

2,2) \(\mathfrak S(R_1R_2)=\mathfrak S(R_1)\mathfrak S(R_2)\);

2,3) \(\mathfrak S(\bar R_1)=\overline{\mathfrak S(R_1)}\).

We shall call a contraction of a word \(S\) any word \(S'\) obtained from \(S\) by deleting some occurrences of letters. From the definition of the set \(\mathfrak S(R)\) the following assertions follow immediately:

У2. If \(S\in\mathfrak S(R)\) and \(S'\) is a contraction of the word \(S\), then \(S'\in\mathfrak S(R)\).

У3. If \(R\) is a simple chain, then \(\mathfrak S(R)\) is the set of all contractions of the word \(R\).

It is not hard to prove also the following assertion:

У4. Suppose there is a formula \(R\simeq R(x_1,\ldots,x_m)^*\), and suppose \(\mathfrak S(R)=\{R_i\}_i\). Denote by \(\hat R\) (respectively \(\hat R_i\)) the value of the function \(R\) (respectively \(R_i\)) on the tuple \(X_1,\ldots,X_m\), where \(X_j\in\mathfrak R'_s\), \(j=1,\ldots,m\). Then

\[ \hat R=\bigvee_i \hat R_i. \]

\(*\) By the sign \(\simeq\) we denote graphical identity.

From U4 the following assertion easily follows:

U5. The identity \(R=S\) is true in \(\mathfrak R'_s\) if and only if
\(\mathfrak S(R)=\mathfrak S(S)\).

Hence it follows immediately:

U6. \(R\subseteq S\) if and only if \(\mathfrak S(R)\subseteq \mathfrak S(S)\).

With the aid of identities 0–4, 8, 10, 11 it is easy to prove the following assertion:

U7. For every formula \(R(x_1,\ldots,x_m)\)
\[ \Sigma' \vdash \overline{R(x_1,\ldots,x_m)}=\overline{x_1\ldots x_m}. \]

From U7 and identities 5, 6 it follows:

U8. For any formula \(R\),
\[ \Sigma' \vdash R=R_1\vee\cdots\vee R_m, \]
where \(R_1,\ldots,R_m\) are chains.

From U3 and U6 it follows:

U9. \(x_{i_1}\ldots x_{i_m}\subseteq x_{j_1}\ldots x_{j_n}\) if and only if the word \(x_{i_1}\ldots x_{i_m}\) is a contraction of the word \(x_{j_1}\ldots x_{j_n}\).

From U9, \(\delta\), 1, 2, 11 it follows:

U10. \(x_{i_1}\ldots x_{i_m}\subseteq \overline{x_{j_1}\ldots x_{j_n}}\) if and only if
\[ \{x_{i_1},\ldots,x_{i_m}\}\subseteq \{x_{j_1},\ldots,x_{j_n}\}. \]

Similarly one obtains:

U11. \(\overline{x_{i_1}\ldots x_{i_m}}\subseteq \overline{x_{j_1}\ldots x_{j_n}}\) if and only if
\[ \{x_{i_1},\ldots,x_{i_m}\}\subseteq \{x_{j_1},\ldots,x_{j_n}\}. \]

From U9–U11, \(9'\), \(9''\) it follows:

U12. If \(R\) and \(S\) are elementary chains and \(R\subseteq S\), then
\[ \Sigma' \vdash R\vee S=S. \]

We shall say that a chain \(R\simeq R_1\ldots R_m\) is a subchain of the chain \(S\simeq S_1\ldots S_n\), where \(R_1,\ldots,R_m,S_1,\ldots,S_n\) are elementary chains, if there exist \(S_{i_1},\ldots,S_{i_m}\) such that
\[ 1\le i_1<i_2<\cdots<i_m\le n \]
and
\[ R_k\subseteq S_{i_k},\qquad k=1,\ldots,m. \]

From U12, \(\gamma\) and 14, 15 it follows:

U13. If \(R\) is a nonempty subchain of the chain \(S\), then
\[ \Sigma' \vdash R\vee S=S. \]

A chain \(R'\) obtained from a chain \(R\) by repeating certain cycles will be called an extension of the chain \(R\).

In view of 8, we have:

U14. If \(R'\) is an extension of the chain \(R\), then
\[ \Sigma' \vdash R'=R. \]

By induction it is not difficult to prove the following assertion.

U15. If \(R\) is a chain, then \(\mathfrak S(R)\) consists of all simple subchains of extensions of the chain \(R\).

From U3 and U6 it follows:

U16. If \(R\) is a simple chain, then \(R\subseteq S\) if and only if
\[ R\in \mathfrak S(S). \]

We shall say that a chain \(R\) is contained in a chain \(S\), if \(R\) is a subchain of some extension of the chain \(S\).

From U13 and U14 it follows:

U17. If a nonempty chain \(R\) is contained in a chain \(S\), then
\[ \Sigma' \vdash R\vee S=S. \]

U18. If \(R\) and \(S\) are chains, then \(R\subseteq S\) if and only if \(R\) is contained in \(S\).

Proof. If \(R\) is contained in \(S\), then \(R\subseteq S\) by U17. Let \(R\subseteq S\); we shall show that then \(R\) is contained in \(S\).

1) Let \(R\) be an elementary chain.

1.1) If \(R\simeq x_i\), the assertion is trivial.

1.2) Let \(R\simeq x_{i_1}\ldots x_{i_m}\). Then: 1,2,1) if in \(S\) there is a cycle
\[ \overline{x_{j_1}\ldots x_{j_n}} \]
such that
\[ \{x_{i_1},\ldots,x_{i_m}\}\subseteq \{x_{j_1},\ldots,x_{j_n}\}, \]
then, obviously, \(R\) is contained in \(S\). 1,2,2) Suppose that for every cycle
\[ \overline{x_{j_1}\ldots x_{j_n}} \]
from \(S\),
\[ \{x_{i_1},\ldots,x_{i_m}\}\nsubseteq \{x_{j_1},\ldots,x_{j_n}\}. \]
Consider the \(R\)-word \(R'\simeq (x_{i_1}\ldots\)

... \(x_{i_m})^\alpha\), where \(\alpha > C\), \(C\) is the “length” of the chain \(S\), i.e., the number of occurrences of variables in \(S\). Let \(S \simeq S_1 \ldots S_l\), where \(S_1,\ldots,S_l\) are elementary chains. According to U15 and U9, U10, every \(S\)-word has the form
\[ S' \simeq (x_{j_1}^{\beta_1}\ldots x_{j_n}^{\beta_n})^{\alpha_1}\ldots \]
\[ \ldots (x_{q_1}^{\gamma_1}\ldots x_{q_k}^{\gamma_k})^{\alpha_p}, \]
where \(p \leq l\) and in the parentheses there stand variables belonging to one cycle from \(S\), or else all exponents of the variables and of the parentheses are equal to 1. Suppose that \(R'\) is a subchain of the \(S\)-word \(S'\). Since no set of variables in parentheses with exponent \(\alpha_s > 1\) contains the set \(\{x_{i_1},\ldots,x_{i_m}\}\), it follows that \(R'\) must be a subchain of an \(S\)-word \(S''\) in which all exponents \(\beta_1,\ldots,\beta_n,\ldots,\gamma_1,\ldots,\gamma_k\) and \(\alpha_1,\ldots,\alpha_p\) are equal to 1. But the length of such a word does not exceed \(C\), which contradicts \(\alpha > C\). Thus, in view of U15, the \(R\)-word \(R'\) is not an \(S\)-word and therefore, by U6, \(R \nsubseteq S\), which contradicts the assumption.

2) Suppose that the assertion is true for \(R'\) and \(R''\), and \(R \simeq R'R''\). Since \(R' \subseteq R'R''\) and \(R'' \subseteq R'R''\), we have \(R' \subseteq S\) and \(R'' \subseteq S\). Let \(S \simeq S_1\ldots S_l\), where \(S_1,\ldots,S_l\) are elementary chains, and let \(p\) and \(q\) be such that
\[ R' \subseteq S' \simeq S_1\ldots S_p,\quad R' \nsubseteq S_1\ldots S_{p-1},\quad R'' \subseteq S'' \simeq S_qS_{q+1}\ldots S_l,\quad R'' \nsubseteq S_{q+1}\ldots S_l. \]
The following cases are possible.

2.1) \(p < q\). Then, since by the induction hypothesis \(R'\) fits into \(S'\), and \(R''\) fits into \(S''\), \(R\) fits into the subchain \(S'S''\) of the chain \(S\), and consequently \(R\) fits into \(S\).

2.2) \(p=q\) and \(S_p\) is a cycle. Then, obviously, \(R\) fits into the expansion
\[ S_1\ldots S_pS_p\ldots S_l \]
of the chain \(S\).

2.3) \(p=q\) and \(S_p\) is a variable. Let \(P \simeq S_1\ldots S_{p-1}\), \(Q \simeq S_{p+1}\ldots S_l\), i.e., \(S \simeq PS_pQ \simeq Px_iQ\). Since \(R' \nsubseteq P\) and \(R'' \nsubseteq Q\), by U6, U15, and U16 there exist an \(R'\)-word \(T_1\) and an \(R''\)-word \(T_2\) such that \(T_1\) is not a subchain of any expansion of the chain \(P\), and \(T_2\) is not a subchain of any expansion of the chain \(Q\). And since every expansion of the chain \(S\) has the form \(P'x_iQ'\), where \(P'\) and \(Q'\) are expansions of the chains \(P\) and \(Q\), the simple chain \(T_1T_2\) is not a subchain of any expansion of the chain \(S\), i.e., by U15, the \(R\)-word \(T_1T_2\) is not an \(S\)-word. In view of U6 this means that \(R \nsubseteq S\), which contradicts the assumption.

2.4) \(p>q\). Then, obviously, there exist chains \(P\) and \(Q\) such that \(S \simeq PQ\) and \(R' \nsubseteq P\), \(R'' \nsubseteq Q\). Analogously to what was done in item 2.3), one can show that then \(R \nsubseteq S\). The assertion is proved.

From U17 and U18 it follows:

U19. If \(R\) and \(S\) are chains and \(R \subseteq S\), then \(\Sigma' \vdash R \vee S = S\).

U20. Let \(R \subseteq S\), where \(S \simeq S_1 \vee \ldots \vee S_n\) and \(R,S_1,\ldots,S_n\) are chains. Then \(\Sigma' \vdash R \vee S = S\).

Proof. In view of U19 and \(\varepsilon\), it is enough for us to prove that if \(R \subseteq S \simeq S_1 \vee \ldots \vee S_n\), then there is an \(S_k\), \(1 \leq k \leq n\), such that \(R \subseteq S_k\). Suppose the contrary, i.e., that \(R \nsubseteq S_k\) for every \(k=1,\ldots,n\). Then, according to U6, for every \(k=1,\ldots,n\) there is an \(R\)-word \(R_k\) that is not an \(S_k\)-word. But in view of U15, for the set of \(R\)-words \(\{R_k\}_{k=1}^n\) there is an \(R\)-word \(R'\) such that every \(R\)-word from this set is its subchain. Since for every \(k=1,\ldots,n\), \(R_k \subseteq R'\) and \(R_k \nsubseteq S_k\), it follows that for every \(k=1,\ldots,n\), \(R' \nsubseteq S_k\). But, obviously, every \(S\)-word is an \(S_k\)-word for some \(k\), \(1 \leq k \leq n\), and therefore \(R' \nsubseteq S\), i.e., in view of U16 \(R \nsubseteq S\). The contradiction obtained proves the assertion.

Obviously:

U21. If \(R_1 \vee \ldots \vee R_m \subseteq S\), then for every \(i=1,\ldots,m\), \(R_i \subseteq S\).

From U8, U20, and U21, in view of \(\zeta\), U1 follows.

Received
30 V 1962

CITED LITERATURE

M. O. Rabin, D. Scott. I. B. M. J. Res. Develop., 3, No. 2, 114 (1959).