Parameterized Regular Expressions and Their Languages

A straightforward idea is to explicitly state even - odd and odd - even cases, that is,redefine e 5 1 as e5 1,e | e5 1,o , wheree 5 1,e = ⋃ a∈Γ∆ ∗ ·x 1···x n ·a·&·(∆\{% even ,% odd }) ∗ ·% even·# odd ·((0|1) n ·(Γ∪(Γ×Q))·& )∗ ·x 1···x n ·((Γ\{a})∪((Γ\{a})×Q) )·∆ ∗e 5 1,o = ⋃ a∈Γ∆ ∗ ·y 1···y n ·a·&·(∆\{% even ,% odd }) ∗ ·% odd·# even ·((0|1) n ·(Γ∪(Γ×Q))·& )∗ ·y 1···y n ·((Γ\{a})∪((Γ\{a})×Q) )·∆ ∗The problem is that these expressions are not simple: they reuse the variables x 1 ,...,x nor y 1 ,...,y n . The solution is instead a bit more technical. We redefine e 5 1 as e6 1,e | e6 1,o , where:e 6 1,e = ⋃ a∈Γ∆ ∗ # even ·(∆\{% even }) ∗·(x 1···x n ·( (a·&·(∆\{%even }) ∗ ·% even ·# odd (∆\{% odd }) ∗ ·& ) |(((Γ\{a})∪((Γ\{a})×Q))·(∆\{%odd ,# even ,% even }) ∗))) ∗·% odd ·∆ ∗e 6 1,o = ⋃ a∈Γ∆ ∗ # odd ·(∆\{% odd }) ∗·(y 1···y n( (a·&·(∆\{%odd }) ∗ ·% odd ·# even (∆\{# even }) ∗ ·& ) |(((Γ\{a})∪((Γ\{a})×Q))·(∆\{%even ,# odd ,% odd }) ∗))) ∗·% even ·∆ ∗Notice then that e 6 1,e | e 6 1,o is a simple parameterized regular expression. In order to seethat the intended meaning of these expressions remains the same, notice that L ✸ (e 5 1,e ) ⊆L ✸ (e 6 1,e) and L ✸ (e 5 1,o) ⊆ L ✸ (e 6 1,o). Moreover, it is not difficult to check that none of thewords that belong to L ✸ (e 6 1,e ) but not to L ✸(e 5 1,e ) represent a valid run of M, and neitherdoes anyword inL ✸ (e 6 1,o ) but not inL ✸(e 5 1,o ). Thus, thewords in L ✸(e 6 1,e ) but not inL ✸(e 5 1,e )(respectively, in L ✸ (e 6 1,o) but not in L ✸ (e 5 1,o)) are not harmful for our purposes, since theseextra words already belong to the language of some other disjunction in e M,ā .With these observations, it is not difficult to modify the remainder of the reduction ofthe proof of Theorem 6 so that every expression is simple. The proof then follows along thesame lines as the proof of Theorem 6.✷32