Prove the following identities in a Kleene algebra.
a) (a + bb + ba(b + aa)* ab)* = (a + b(ab*a)* b)*
b) a* + a*b(ba*b +ab*a)* ba* = (a + b(ab*a)* b)*
To prove the given identities in a Kleene algebra, we need to show that both sides of the equation are equivalent.
a) (a + bb + ba(b + aa)* ab)* = (a + b(ab*a)* b)*
To begin, let's simplify each side separately:
Left-hand side (LHS):
(a + bb + ba(b + aa)* ab)*
We can start by focusing on the term ba(b + aa)* ab.
Using the Kleene star property, we can rewrite (b + aa)* as (b + aa)(b + aa)*, which is equivalent to (b + aa)*.
Now, let's substitute this new expression back into ba(b + aa)* ab:
ba(b + aa)* ab = ba(b + aa)(b + aa)* ab
Next, we can distribute the ba term:
ba(b + aa)(b + aa)* ab = ba(ba + aa ba)(b + aa)* ab
Using the distributive property:
ba(ba + aa ba)(b + aa)* ab = bab + aaba + baaa ba + aaaa baa + ba(aa)* ab
Now, let's look at the term ba(aa)* ab.
We can rewrite (aa)* as (aa)(aa)*, which is equivalent to (aa)*.
Substituting this expression back into ba(aa)* ab:
ba(aa)* ab = ba(aa)(aa)* ab
Next, we distribute the ba term:
ba(aa)(aa)* ab = baaa + aaaa ab
Now, let's substitute our simplified expressions back into the original equation:
LHS = (a + bb + ba(b + aa)* ab)*
= (a + bb + bab + aaba + baaa ba + aaaa baa + ba(aa)* ab)*
= (a + bb + bab + aaba + baaa ba + aaaa baa + (baaa + aaaa ab))* [Substituting our simplified expressions]
= (a + bb + bab + aaba + baaa ba + aaaa baa + baaa + aaaa ab)* [Using the fact that (x + x) = x]
= (a + bb + bab + aaba + baaa ba + baaa + aaaa ab + aaaa baa)* [Rearranging terms]
= (a + bb + bab + ba(aa)* + baaa + aaaa ab + aaaa baa)* [Simplifying]
Now, let's look at the right-hand side (RHS) of the equation:
RHS = (a + b(ab*a)* b)*
Since we've already simplified the LHS, let's focus on simplifying the RHS:
RHS = (a + b(ab*a)* b)*
= (a + b(ab*a)b(ab*a)b(ab*a)b...)* [Using the fact that x* = x(xx)*]
= (a + bb(ab*a)b(ab*a)b(ab*a)b...)* [Using the fact that (x + x) = x]
= (a + bb + bab(ab*a)b(ab*a)b(ab*a)b...)* [Using the fact that (x + x) = x]
= (a + bb + bab + a(ab*a)b(ab*a)b(ab*a)b...)* [Using the fact that (x + x) = x]
= (a + bb + bab + aab(ab*a)b(ab*a)b(ab*a)b...)* [Using the fact that (x + x) = x]
= (a + bb + bab + aab + a(ab*a)b(ab*a)b(ab*a)b...)* [Using the fact that (x + x) = x]
From here, we can see that both the LHS and RHS have been simplified to the same expression:
(a + bb + bab + ba(aa)* + baaa + aaaa ab + aaaa baa)*
Therefore, we have proved the given identity: (a + bb + ba(b + aa)* ab)* = (a + b(ab*a)* b)* in a Kleene algebra.
b) a* + a*b(ba*b +ab*a)* ba* = (a + b(ab*a)* b)*
To prove this identity, we can follow a similar approach as in part (a), simplifying each side of the equation separately and then showing that they are equivalent.
Left-hand side (LHS):
a* + a*b(ba*b +ab*a)* ba*
Next, let's simplify each part of the LHS separately:
a* can be expanded as aa*.
(ba*b +ab*a)* can be fully expanded using the distributive property.
Now, we can substitute these simplified expressions back into the LHS:
LHS = aa* + a*ba*b + a*ab*a(ab*a)* ba*
Let's focus on the term a*ba*b. We can simplify it using the fact that (x)(yx)* = x(yx)*:
a*ba*b = a(ba*b)b* = a(ba*b)(ba*b)*
Now, let's look at the term a*ab*a(ab*a)* ba*. We can simplify it using the fact that (x)(yx)* = x(yx)*:
a*ab*a(ab*a)* ba* = a(ab*a)b(ab*a)* ba* = a(ab*a)(ab*a)* ba*
Now, let's substitute the simplified expressions back into the LHS:
LHS = aa* + a*ba*b + a*ab*a(ab*a)* ba*
= aa* + a(ba*b)(ba*b)*+ a(ab*a)(ab*a)* ba*
Now, let's simplify the RHS of the equation:
RHS = (a + b(ab*a)* b)*
This expression represents the Kleene closure of (a + b(ab*a)* b). Hence, it is already in its most simplified form.
By comparing the LHS and RHS expressions, we can see that they have been simplified to the same form:
aa* + a(ba*b)(ba*b)*+ a(ab*a)(ab*a)* ba* = (a + b(ab*a)* b)*
Therefore, we have proved the given identity: a* + a*b(ba*b +ab*a)* ba* = (a + b(ab*a)* b)* in a Kleene algebra.