; XOR-Form : The statement also holds when the
disjunction "+" is replaced by the
XOR operator: : f(X_1, X_2, \dots , X_n) = X_1 \cdot f(1, X_2, \dots , X_n) \oplus X_1' \cdot f(0, X_2, \dots , X_n) ; Dual form: There is a dual form of the Shannon expansion (which does not have a related XOR form): : f(X_1, X_2, \dots , X_n) = (X_1 + f(0, X_2, \dots , X_n)) \cdot (X_1' + f(1, X_2, \dots , X_n)) Repeated application for each argument leads to the
Sum of Products (SoP) canonical form of the Boolean function f. For example for n=2 that would be :\begin{align} f(X_1, X_2) & = X_1 \cdot f(1, X_2) + X_1' \cdot f(0, X_2)\\ & = X_1 X_2 \cdot f(1, 1) + X_1 X_2' \cdot f(1, 0) + X_1' X_2 \cdot f(0, 1) + X_1' X_2' \cdot f(0, 0) \end{align} Likewise, application of the dual form leads to the
Product of Sums (PoS) canonical form (using the
distributivity law of + over \cdot): :\begin{align} f(X_1, X_2) & = (X_1 + f(0, X_2)) \cdot (X_1' + f(1, X_2))\\ & = (X_1 + X_2 + f(0, 0)) \cdot (X_1 + X_2' + f(0, 1)) \cdot (X_1' + X_2 + f(1, 0)) \cdot (X_1' + X_2' + f(1, 1)) \end{align} == Properties of cofactors ==