# Proving language is not context free using pumping lemma

Hi I'm completely stuck on an exercise which is to prove this language is not context free using pumping lemma for context free languages:

L = {xyz | x + y = z} , where the alphabet is from 0-9, so for example
10^m20^m30^m ∈ L for all m ≥ 0 (since 1+2 =3, 10 + 20 = 30, etc)



How would you solve this using the pumping lemma for context free languages? Thank you for your time :)

• Try to imitate the proof given here that $a^nb^nc^n$ is not context-free. – saulspatz May 18 at 20:56
• I have tried to replicate already and was looking at this video: youtube.com/watch?v=AdfE0IcGaJs&t=57s , but I can't figure out how it connects to amount of 0s etc and I imagine it's not exactly the same. But thank you anyway. @saulspatz – Blue shirt May 18 at 21:05
• Simple question. When dividing string = uvxyz, can |u|=0? I'm wondering because if that the case then vxy could contain 10i where i>=p, but I'm not sure. Thank you! @saulspatz – Blue shirt May 18 at 21:33
• According to Wikipedia, yes $|u|=0$ is allowed. I can't pretend that I remember all the technicalities of the definition myself. – saulspatz May 18 at 23:25