For symmetric $M=\left(\begin{array}{cc}A& B\\ B^{\mathsf{T}}& C\end{array}\right)$ with $A$ invertible, a block-unitriangular congruence gives $A\oplus (C-B^{\mathsf{T}}{A}^{-1}B)$ and factors $\det M$

If the total size is zero, every block and $P$ is empty, both displayed matrix identities are the empty identity, and the determinant formula reads $1=1\cdot 1$. Otherwise $P$ is block upper triangular with identity diagonal blocks, and $P^{-1}=\left(\begin{array}{cc}I& A^{-1}B\\ 0& I\end{array}\right)$; hence it is invertible and [L4] gives $\det P=1=\det P^{\mathsf{T}}$.

Since symmetry of $M$ makes $A^{\mathsf{T}}=A$ and therefore $(A^{-1}{)}^{\mathsf{T}}=A^{-1}$, direct multiplication gives $P^{\mathsf{T}}MP=\left(\begin{array}{cc}A& 0\\ 0& C-B^{\mathsf{T}}{A}^{-1}B\end{array}\right)=A\oplus S$.

In positive total size, [L3] and step 1.1 give $\det (P^{\mathsf{T}}MP)=\det M$. In the Leibniz sum [L2] for the block diagonal matrix of step 2.1, every nonzero term preserves both index blocks, so its determinant is $\det A\det S$, with the stated empty-block convention when one block has size zero. Thus $\det M=\det A\det S$.

The calculation proves both assertions. Zero-sized blocks, zero entries inside $A,B,C$, and singular $S$ require no cancellation and are all included.