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For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable.
"Positiveness Problem".
Note that each instance of the Positiveness Problem is a regular language. (Each instance itself is, not the set of solved instances.) Clearly, we cannot take the set of strings in the language to be our instance, (since, in general, this is likely to be in?nite in size. But we have at least two means of specifying any regular language using ?nite objects: we can give a Finite State Automaton that recognizes the language as a ?ve-tuple, each component of which is ?nite, (or, equivalently, the transition graph in some other form) or we can give a regular expression. Since we have algorithms for converting back and forth between these two forms, we can choose whichever is convenient for us. In this case, lets assume we are given the ?ve-tuple. Since we have an algorithm for converting NFAs to DFAs as well, we can also assume, without loss of generality, that the automaton is a DFA.
A solution to the Positiveness Problem is just "True" or "False". It is a decision problem a problem of deciding whether the given instance exhibits a particular property. (We are familiar with this sort of problem. They are just our "checking problems"-all our automata are models of algorithms for decision problems.) So the Positiveness Problem, then, is just the problem of identifying the set of Finite State Automata that do not accept the empty string. Note that we are not asking if this set is regular, although we could. (What do you think the answer would be?) We are asking if there is any algorithm at all for solving it.
Given any NFA A, we will construct a regular expression denoting L(A) by means of an expression graph, a generalization of NFA transition graphs in which the edges are labeled with
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Exercise Show, using Suffix Substitution Closure, that L 3 . L 3 ∈ SL 2 . Explain how it can be the case that L 3 . L 3 ∈ SL 2 , while L 3 . L 3 ⊆ L + 3 and L + 3 ∈ SL
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Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con
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