Question

Is the following regular expression equivalence true? Why or why not?

(ab)* u (aba)* = (ab u aba)*

*=Kleene star

u=Union (Set Theory)

Answer 1

No, they aren’t equivalent. The language on the RHS contains “abaab”, but the language on the LHS doesn’t. Is there any relationship among these? Yes; but I won’t just give you the answer. Hint: are there any strings in the RHS that aren’t in the LHS?

EDIT:

Just to expound a little for the interested reader. Languages are sets of strings. Therefore, relationships among sets are also relationships among languages. The most common set relationships are equality, subset, and superset. Given two sets A and B over universe sets U1 and U2, set A is a subset of B under universe set U1 u U2 (u stands for union) iff every element of A is also an element of B. Similarly, given two sets A and B over universe sets U1 and U2, set A is a superset of B under universe set U1 u U2 (u stands for union) iff every element of B is also an element of A (equivalently, iff B is a subset of A). Sets A and B are equal iff A is a subset of B and B is a subset of A (equivalently, iff A is a superset of B and B is a superset of A). Note that two sets A and B need not be in any of these three relationships; this happens when A contains an element not in B and, at the same time, B contains an element not in A.

To find which of the four possible relationships exist between sets A and B – equality, subset, superset, none – you usually check whether A is a subset of B and whether B is a subset of A. Call the first check B1 and the second check B2, where B1 and B2 are Boolean variables and are true iff the checks pass (i.e., A is a subset of B in the case of B1, and B is a subset of A in the case of B2). Then (B1 && B2) corresponds to equality, (B1 && !B2) corresponds to subset, (!B1 && B2) corresponds to superset and (!B1 && !B2) corresponds to no relationship.

In the example above, I demonstrate that the two languages are not equal by demonstrating that the RHS contains an element not in the LHS. Note that this also rules out the “A is a superset of B” relationship. Two remain: B is a superset of A, or there is no relationship between them. To decide this, you must determine whether A is a subset of B; whether the elements in the LHS language are all in the RHS language. If so, the LHS is a subset; otherwise, there is no relationship.

To show that there is an element in one set and not another, the easiest and most convincing approach is a proof by counterexample: name a string in one but not the other. This is the approach I adopted. You can also make an argument that the language must contain such an element without explicitly naming it; this kind of proof can be significantly harder to get right.

To show that every element of set A is also in set B, you will need a more generic proof technique. Proof by induction and proof by contradiction are common examples. To do an inductive proof, you assign – explicitly or implicitly – a natural number to each string in the language, demonstrate that your claim (in this case, that the element is also an element of the other set) is true with a simple argument. Then, you assume it is true for the first n elements in your set (according to the numbering you give) and show that this implies all the elements that come afterwards must also satisfy your claim. To do a proof by contradiction, you assume the opposite of what you want to prove, and derive a contradiction. If you succeed and your only assumption was that your claim is false, then your claim must have been true all along.