${\mathbf{M}}_B$ from ${\mathbf{M}}$ to leave a valid subnormalised ${\mathbf{M}}' = {\mathbf{M}}-{\mathbf{M}}_B$. In this case at least one of the following must also be true:

    1.  There exists an $x^*$, and a set $y_a$, such that $M(ab|x^* y_a)>0 \; \forall \, a,b$. It is possible to subtract the subnormalised basic measurement $M_B(ab|xy)=k \delta_{x, x^*}\delta_{y, y_a}$ from ${\mathbf{M}}$, where $k=\min_{a,b} M(ab|x^* y_a)$, generating at least one additional 0 element in ${\mathbf{M}}'$. \[wiring1\]

    2.  There exists a $y^*$, and a set $x_b$, such that $M(ab|x_b y^*)>0 \; \forall \, a,b$. It is possible to subtract the subnormalised basic measurement $M_B(ab|xy)=k \delta_{y, y^*}\delta_{x, x_b}$ from ${\mathbf{M}}$, where $k=\min_{a,b} M(ab|x_b y^*)$, generating at least one additional 0 element in ${\mathbf{M}}'$. \[wiring2\]

2.  It is not possible to subtract a (non-zero) subnormalised basic measurement from ${\mathbf{M}}$ without creating a negative component. In this case there exist sets $a_x$, and $b_{xy}$, such that $M(a_x b_{xy}  |xy)=0 \; \forall \, x,y$, and there exist sets $b_y$, and $a_{xy}$, such that $M(a_{xy} b_y  |xy)=0 \; \forall \, x$ \[nowiring2\]. In terms of the representation of ${\mathbf{M}}$ as an array, this means that each row of blocks (corresponding to fixed $x$) contains a row of components (corresponding to fixed $x$ 