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.