Can someone help me with this proof using the pumping lemma?

Question

I just started reading about the pumping lemma and know how to perform a few proofs, mostly by contradiction. It is only this particular question which I don't seem to find an answer for. I have no idea on how to begin. I can assume that there has to be a pumping length P and that for all w element of L that the LENGTH(w) >= P. And of course that we can write w as xyz with the three normal conditions of the pumping lemma.

I have to proof that the following language is non regular:

L = {x + y = z | x,y,z element of {0,1}* and #(x) + #(y) = #(z) }

Can someone help me on this, I really want to master the process in proofing these kind of questions?

Edit:
Sorry, forgot to say that the alphabet is {0,1,+,=} and # means the binary value of the string. Like #(00101) = 5 and #(110) = 6.

Answer 1

Since you want to master the process, I'll point out a few things before showing a proof.

1. The first thing to notice is that the + and the = may only appear once each. So when you write your string w as w = abc, the pumped portion, b, cannot contain + or = otherwise you'd reach a trivial contradiction (I'm not using the more standard w = xyz notation to avoid confusion with L's definition).

2. Another thing to notice is that normally, you'd pick a specific string w to pump. In this case, it could be easier to pick a class of strings that share a certain property. The pumping lemma only requires you to reach a contratiction using one string, but there's no reason you can't reach a contradiction with multiple strings.

Proof (in a spoiler):

So let w be any string in L such that |w| ≥ P and x, y, z do not contain leading 0's. By the pumping lemma we can write w as w = abc By pumping lemma, we know b is not empty. Since b cannot contain + or =, it is fully contained in either x, y, or z. Pumping w with any i ≠ 1 results in the binary equation no longer holding since exactly one of x, y, z would be a different number (this is why we needed the no leading 0's bit).

Answer 2

Choose as the string 1(0^n+1) + 1(0^n) = 11(0^n).

In other words, your string will read "the sum of two to the power n+2 plus two to the power n+1 is equal to 11 followed by n zeroes".

Since the string to be pumped will consist entirely of symbols from the first addend, pumping must change the number represented (adding or removing digits to a number will change the number; this is true because our string doesn't contain leading zeroes) and if x + y = z holds, then x' + y = z does not hold if x' != x (over integers, at least).

Since the pumping lemma requires pumped strings to be in the language, and pumping this string fails, we have that the language is not regular.