= SOP element-state conservation
{c}
{title2=$p_I+p_1+p_2=1$}
Represent a three-state <probability distribution> by occupancy fractions $\boldsymbol p=(p_I,p_1,p_2)$ in $I,A_1,A_2$. A nonnegative transfer $f_{ij}\leq p_i$ subtracts the same amount from donor $i$ that it adds to recipient $j$. The total therefore remains $p_I+p_1+p_2=1$. If all simultaneous outflows from a state sum to at most its old occupancy, each occupancy remains nonnegative. This elementary conservation argument makes a proposed numerical realization checkable; it does not identify the states with biological compartments.
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