For each explain your answer a ba ab b ab ba bb aa 0 c For

For each, explain your answer. a). b*a* (ab)* b). a*b* b*a* b*b* a*a* = 0 c). For x, y 1*, xy = yx iff | sigma | = 1. d). For x, y, z Z*, (xy)z = x(yz) iff | sigma | = 1. e). (a* U b*)* can be reduced to sigma* iff sigma = {a, b} f). If sigma is any alphabet, and e L_1 and e L_2, then (L_1 sigma* L_2) = sigma^+ g). |L_1L_2| = |L_1||L_2|

Solution

a) False

b*a* = b, a, ba, bba, baa, bbaa,...

(ab)* = ab, abab, ababab,...

b) True

There is no such element which is common in all of these expressions.

a*b* = a, b, ab, aab, abb, ...

b*a* = b, a, ba, bba, baa, ...

b*b* = b, bb, bbb, ...

a*a* = a, aa, aaa, ...

c) True

If the input set has only one element then the rule of commutativity will hold.

d) True

If the input set has only one element then the rule of associativity will hold.