Prove L = {a = b ⊕ c | a, b, c ∈{0, 1}*} is not regular

Question

Given that language $L = \{a = b ⊕ c \mid a, b, c ∈ \{0,1\}^*, a = b \oplus c\}$, with an alphabet $Σ = \{⊕, =, 0, 1\}$, I need to prove that this language is not regular. The following is as far as I have gotten in my proof, but I am getting stuck when showing why it is contradicting the pumping lemma:

$s = (1^p = 1^p ⊕ 0^p)$

$x = 1^l$

$y = 1^m$, where $l+m \le p$ and $m > 0$

$z = (1^{p-l-m}=1^p ⊕ 0^p)$

Taking $i=2$, $xy^2z = (1^l1^{2m}1^{p-l-m}=1^p ⊕ 0^p) = (1^{p+m}=1^p ⊕ 0^p)$

—This is where I am getting stuck. From my example, this appears to be a contradiction, but there is nothing in the language dictating that $b$ and $c$ be of equal lengths. Were they to be different lengths, it could certainly be possible that no contradiction would be found by adding $m$ to $p$. How could I fix my proof via pumping lemma?

---EDIT: SECOND TRY---

$s = ((10)^p = (01)^p ⊕ (10)^p)$

$x = (10)^l$

$y = (10)^m$, where $l+m \le p$ and $m > 0$

$z = ((10)^{p-l-m}=(01)^p ⊕ (10)^p)$

Taking $i=0$, $xy^0z = ((10)^l(10)^{p-l-m}=(01)^p ⊕ (10)^p) = ((10)^{p-m}=(01)^p ⊕ (10)^p)$

Now, I have that $|a|<|b|$ and $|a|<c$, which is not possible, since the result of the exclusive-or must contain a leading 1, which cannot be truncated to a smaller string length. Therefore, this is a contradiction.

Is this a contradiction strong enough to prove the language is not regular? It only works in the case of a leading 1; do I need to show something that is a contradiction even with leading 0's?

Answer 1

Suppose $L$ were regular. Let $p$ be the pumping constant assumed to exist and consider the string

$$s = 1^p = 1^p \oplus 0^p$$

Clearly, $|s| \geq p$ so for some strings $x,y,z$ we have $s = xyz$ where

1. $|xy| \leq p$
2. $|y| > 0$
3. $xy^iz \in L$ for all $i \geq 0$

If condition 1 is to hold, $y$ must be a string of $1$s. If condition 2 is to also hold, then $y = 1^k$ for some $k > 0$. But then condition 3 fails, since $xy^0z \notin L$.

For $xy^0z = 1^{p-k} = 1^p \oplus 0^k$, and this identity is false irrespective of whether we require $w_1 \oplus w_2$ to be defined for strings of the same length only, or in general.

Answer 2

Thought A)

Perhaps your definition of "a" needs to be reworked; eg. not count all the leading zeros as significant (0000 = 0). You can substantiate this by making the case that XOR is an arithmetic operation.

In this way given an acceptor for a given length of string you could pump through another string that output the same result proving that its not regular.

010 XOR 010 = 0 and 0110 XOR 0110 = 0

Thought B)

There are no "real" contradictions for the language "a". It consists of every string of { 0,1 }* by definition. However using thought A) you can pump through multiple "definitions" of an accepted word in the language.

Thought C)

What do you think?