Prove or disprove that the following language is context-free
You can use the pumping lemma in this way.

By closure properties, if $L$ is context free and it is intersected with the regular language $R = \{ 1^* 0^* 2 1^* 0^*$} the resulting langauge $L' = L \cap R$ is still context free.

We can apply the Pumping lemma to $L'$ and prove that it cannot be CF, so $L$ is not CF as well. 

Let $p$ be the pumping length; pick

$$z = 1^p \, 0 \, 0^p \; 2 \; 1^{p} \, 1 \, 0 ^p$$

First note that if $vwx$ is entirely contained in the first half, then pumping it one time is enough to make the first half longer than the second half; making $[\alpha]_2 &gt; [\beta]_2$ and pushing the string out of $L$

If $vwx$ is entirely contained in the second half, then pumping it zero times is again enough to make the first half longer than the second half; making $[\alpha]_2 &gt; [\beta]_2$ and pushing the string out of $L$

So $vwx$ crosses the middle of $z$.

At least one, but at most $p$ positions are contained in $vwx$, so $v$ must be in the $0^p$ before the center and $x$ in the $1^p$ after the center.

If $|v| &lt;&gt; |x|$ then - as above - pumping them one (or zero times) would make the first half longer.

But if $|v| = |x| = k \geq 1$ we can still pump it zero times getting the string:

$$z^0 = 1^p \, 0 \, 0^{p-k} \; 2 \; 1^{p-k} \, 1 \, 0^p$$

but 

$$ [ 1^p \, 0 \, 0^{p-k} ]_2 &gt;= [ 1^{p-k} \, 1 \, 0^p ]_2$$

(for $k=1$ we have $1^p 0 0^{p-1} = 1^{p-1} 1 0^p$)