SOP element-state conservation ($p_I+p_1+p_2=1$)

Represent a three-state probability distribution by occupancy fractions $\mathbf{p}=(p_I,p_1,p_2)$ in $I,A_1,A_2$. A nonnegative transfer $f_{ij}\le p_i$ 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.