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By Julius T. Tou (Eds.)

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As far as I know, no one has put forth a set of rules that are valid, simple, formally precise, and complete. ) In my opinion, this open problem is worth looking into. 9. STAR HEIGHT These, then, are the four methods of establishing the validity of regularexpression equations. This paper will conclude with a discussion of the second problem, namely, given a regular expression, what interesting or AN INTRODUCTION TO REGULAR EXPRESSIONS 53 (in some significant sense) more simple regular expression is it equal to?

Earlier examples (Sects. 35) of combining partitions are valid here also. 27 Theorem. The algebra (5 of the set of all congruences on an algebra O, the operations being ΓΛ and U as just defined, is a lattice. The lattice is complete [Birkhoff]. It is not hard to show that, for all Ä, S, T £ C, Ä Π Ä = Ä, Ä H S = SHÄ, Ä n (sn r) = ( Ä n s) n r, R Π (Ä U 5) = Ä, R\J R = Ä, R\J S = RV S\JR, (SVJT) = (R\J s)u τ, SU(inS)=fi, so that, by Sect. 22, e is a lattice. Completeness is harder to prove. 28 The theory of congruences and the associated quotient structures is basic in the study of algebraic automata and their decomposition into smaller machines.

Replacement: From an equation a = β, and from an equation in which a occurs as a well-formed part, the result of replacing an occurrence of a by β may be obtained. Example: from (x*i/)* = (x*y)*\J yyy and (x*y) ** = (x*y) *, infer (x*y) * = (x*y) ** VJ yyy. 42 ROBERT MCNAUGHTON Star introduction: From a = aß\J 7, derive a = 7/3*, provided λ ζβ. , see Salomaa [4], p. ) Example: take a = (0 \J 1)*1 \J λ, β = 0*1, and 7 = λ. Then from (0 U 1)*1 U λ = [(0 U 1 ) * 1 U λ]0*1 VJ λ, one can infer ( 0 U 1 ) * 1 U Â = λ(0*1)*, since in this case λ $ β.

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