Theorem: The language $C = \setbuilder{x\#y}{x, y \in \set{0,1}^* \tand x \neq y}$ is context free.

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Proof: Note that $z \in C$ if and only if $z$ is of the form $x \# y$ and one of the following is true: (1) $|x| < |y|$, (2) $|y| < |x|$, or (3) $|x| = |y|$ and there is some $n$ such that $x_n \neq y_n$.

Let the stack alphabet be $\set{k, Z}$ with bottom stack symbol $Z$, and let $x$ represent an arbitrary member of $\set{0, 1}$. We design a PDA as follows:

First, the machine progresses through the substring $x$, pushing a $k$ onto the stack each time it reads a character. Note that, after reading $n$ characters, the machine is in $q_0$ or $q_1$ if the $n$'th character was $0$ or $1$ respectively, and has $n$ instances of $k$ on the stack.

The bottom path $q_{u1} \rightarrow q_e \rightarrow q_{u2}$ (for 'unequal', 'equal', and 'unequal', respectively) handles length comparison. After the $\#$ is read, it immediately accepts until the stack is emptied, at which point the strings are equal length, at which case the machine is in $q_e$. Here, we do not accept on length comparison, and instead rely on the machine to be in $q_a$ to accept (explained below). If any more input is received, the machine goes to $q_{u2}$ and is permanently accepting again. In addition, if a $\#$ is read as the first character, the two strings are equal length because they are both empty, so the start state transitions to $q_e$.

The other path handles comparison on equal-length strings. Every time a character is read, the machine 'spawns a thread' that immediately progresses to $q_{0i}$ or $q_{1i}$. The 'spawned thread' has knowledge of what the $n$'th character of $x$ was, along with $n$ instances of $k$ on the stack. This thread then ignores the rest of $x$, and waits for a $\#$ symbol. Finally, each version of the machine that makes it to $q_{0c}$ or $q_{1c}$ progresses to the $n$'th character of $y$ (by popping the $n$ instances of $k$ off the stack) and tests if the $n$'th character of $y$ is different than what it knew was the $n$'th character of $x$. If this test ever passes, a version of the machine enters $q_a$ and stays there forever, ensuring that the string is accepted.

This machine accepts $C$, so $C$ is context-free.