Similarly, $M(a_{xy} b  |x y)=0 \; \forall \, b,x, y$. In terms of the array, each subarray, corresponding to particular values of $x$ and $y$, contains both a row and a column of zeroes.

But now consider an arbitrary element $M(a^* b^*| x^* y^*)$. Let ${\mathbf{P}}$ be a bipartite product state with $P(a^* b^*| x^* y^*)=1$ and $P(a^* b_{x^* y} |x^* y)=1 \; \forall \, y \neq y^*$, and ${\mathbf{P}}'$ a bipartite product state obtained from ${\mathbf{P}}$ by swapping $a=a^*$ with $a=a_{x^* y^*}$ when $x=x^*$. From Equation (\[diff\_eqn\]), $M(a^* b^*| x^* y^*) = 0$. This contradicts the initial assumption that ${\mathbf{M}}$ is nonzero. Hence case \[nowiring2\] is impossible, and either case \[wiring1\] or case \[wiring2\] must hold. Iteratively subtracting subnormalised basic measurements from ${\mathbf{M}}$ until the zero vector remains yields an expression for ${\mathbf{M}}$ as a convex combination of basic measurements.

We now illustrate the above proof for an example bi-partite measurement, where each subsystem is characterised by two fiducial measurements each with three outputs. Consider a situation in which case \[nowiring2\] holds (it is not possible to remove a basic measurement), and the representation of ${\mathbf{M}}$ as an array contains zeroes as follows (with all other elements unknown): $$\left( 