Looking for a particular algebraic mapping from one Boolean matrix to another Consider the following Boolean matrix:
\begin{align}
X&=\begin{bmatrix}
1&1&1&1&1&1&1&1\\
1&1&1&1&0&0&0&0\\
0&0&0&0&1&1&1&1\\
1&1&0&0&1&1&0&0\\
0&0&1&1&0&0&1&1\\
\end{bmatrix}.
\end{align}
Computing $XX^\top$, we have
\begin{align}
XX^\top&=\begin{bmatrix}
8&4&4&4&4\\
4&4&0&2&2\\
4&0&4&2&2\\
4&2&2&4&0\\
4&2&2&0&4
\end{bmatrix}.
\end{align}
I'd like to produce a new Boolean matrix $Y$ that sets the nonzero elements of $XX^\top$ to zero and the zero elements to one, i.e.,
\begin{align}
Y&=\begin{bmatrix}
0&0&0&0&0\\
0&0&1&0&0\\
0&1&0&0&0\\
0&0&0&0&1\\
0&0&0&1&0
\end{bmatrix}.
\end{align}
Is there a standard algebraic mapping that takes a Boolean matrix $X\in\{0,1\}^{M\times N}$ and returns the desired Boolean matrix $Y\in\{0,1\}^{M\times M}$? My intuition tells me to look into the Kronecker delta as a means of achieving this, but I'm wondering if there is a simpler alternative.

You can simply perform the product $XX^\top$ in the Boolean semiring (which amounts to use the rule $1+1 = 1$ instead of $1 + 1 = 2$). Then just change every $1$ by $0$ and every $0$ by $1$ in the resulting matrix.
