My problem is the following;
Prove that L = $ \{wuv \mid w + u = v\}$ is not context free using the pumping lemma. For instance, the string 12719 ∈ L since 12 + 7 = 19, and the string $10^n20^n30^n ∈ L$ for all $n ≥ 0$ (since $1 + 2 = 3, 10 + 20 = 30, 100 + 200 = 300$ etc.).
To do this I had the idea of using the length of the strings in L. The number of digits of a sum is always equals to or greater then the digits of the addends. For example $300 + 900 = 1200$ is 3 digits + 3 digits = 4 digits. I came up with this language:
L2 = $ \{w^au^bv^c \mid a\leq c$ and $ b < c$ or $b\leq c$ and $ a < c \}$ where $w$ acts like the first integer, $u$ the second integer, and $v$ the sum.
Because L $\subset$ L2, does that mean that if i prove L2 to be not context free, L is not aswell?
Am I on the right track or totally way off? If Im way off, can you point me in the right direction?
A subset of a non-context-free language can be context-free, so your approach with L2 seems off-track, although observations about the numbers' lengths are useful.
The $10^n20^n30^n$ example hints at a strategy: Take $n$ large enough so that the pumping lemma applies to $s$ = $10^n10^n20^n$ (or maybe some other string in $L$ like $1^n2^n3^n$ or $1^n1^n2^n$), then show that no matter how the pumping lemma subdivides $s$, it produces at least one expanded version of $s$ that is not in $L$. This is similar to the example here but probably requires more case analysis to rule out the possible locations of the boundaries of $w,u,v$ (this is where the length observations will help). You might be able to simplify this using the fact that every number is congruent mod 9 to the sum of its digits.