How to use the Pumping Lemma to prove that a restricted subset of 0*1*2*3*, where there are as many 3's as 0's and 1's, is not a CFL?

Question

Use the pumping lemma for context-free languages to show that the following language is not context-free: $ L = \{0^i 1^j 2^i 3^k \mid k=i+j \} $

So I have started like this:

Let us assume that $ L = \{0^i 1^j 2^i 3^k \mid k=i+j \} $ is a context-free language. Then $ ∃ p∈N $ such that $∀ w∈L$, $|w|≥p$, then $∃ u,v,x,y,z∈Σ^*$ such that we can write $w=xuyvz$ which is subject to the following conditions:

$ |uyv|≤p\\ |uv|>0\\ ∀ k≥0,xu^k yv^k z∈L $

Now, let $p$ be the constant in the Pumping lemma and let $w=0^q 1^p 2^q 3^{p+q}$.

Here I have gotten stuck. I am not sure about my choice of $w$ and very unsure about what I can actually say about $w$ since I know nothing about $q$ compared to $p$ except that the number of $3$'s is $p+q$. I am even doubting that it is possible to show that $L$ is not context-free. At the moment there is always the possibility that $uyv$ is in both the $1$'s and $3$'s or the $0$'s and $2$'s (if I swap $p$ and $q$), and then I could possibly pump them just fine.

What am I missing here? Been starring myself blind trying to find it.

Answer 1

Use $|uyv|\le p$ to avoid the cases you mention, e.g., that $u,v$ pump both the $1$'s and the $3$'s. You can choose $p,q$ in $w$ such a way that this way of pumping is not possible. There is no problem in choosing $p=q$!