Boolean Algebra and Simplification Techniques
195
Y
Z
X
X+(Y+Z)
Y
Z
X
Z+(X+Y)
(a)
Y
Z
X
X.(Y.Z)
X
Y
Z
(X.Y).Z
(b)
Figure 6.1 Associative laws.
6.3.7 Theorem 7 (Distributive Laws)
(a) XXXY + ZZ = XXY + XXZ and (b) X + YYZ = X + YYYYX + ZZ
(6.17)
Theorem 7(b) is the dual of theorem 7(a). The distribution law implies that a Boolean expression can
always be expanded term by term. Also, in the case of the expression being the sum of two or more
than two terms having a common variable, the common variable can be taken as common as in the case
of ordinary algebra. Table 6.1 gives the proof of theorem 7(a) using the method of perfect induction.
Theorem 7(b) is the dual of theorem 7(a) and therefore its proof is implied. Theorems 7(a) and (b) are
further illustrated by the logic diagrams in Figs 6.2(a) and (b). As an illustration, theorem 7(a) can be
used to simplify AAB + AAB + AAB + AAB as follows:
AAB + AAB + AAB + AAB = AAAB + BB + AAAB + BB = AA1 + AA1 = A + A = 1
Table 6.1 Proof of distributive law.
X
Y
Z
Y +Z
X Y
X Z
X ( Y +Z)
XY+XZ
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
1
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
1
1
0
1
1
1
1
1
0
1
1
0
1
1
1
1
1
1
1
1
1
1
195
Y
Z
X
X+(Y+Z)
Y
Z
X
Z+(X+Y)
(a)
Y
Z
X
X.(Y.Z)
X
Y
Z
(X.Y).Z
(b)
Figure 6.1 Associative laws.
6.3.7 Theorem 7 (Distributive Laws)
(a) XXXY + ZZ = XXY + XXZ and (b) X + YYZ = X + YYYYX + ZZ
(6.17)
Theorem 7(b) is the dual of theorem 7(a). The distribution law implies that a Boolean expression can
always be expanded term by term. Also, in the case of the expression being the sum of two or more
than two terms having a common variable, the common variable can be taken as common as in the case
of ordinary algebra. Table 6.1 gives the proof of theorem 7(a) using the method of perfect induction.
Theorem 7(b) is the dual of theorem 7(a) and therefore its proof is implied. Theorems 7(a) and (b) are
further illustrated by the logic diagrams in Figs 6.2(a) and (b). As an illustration, theorem 7(a) can be
used to simplify AAB + AAB + AAB + AAB as follows:
AAB + AAB + AAB + AAB = AAAB + BB + AAAB + BB = AA1 + AA1 = A + A = 1
Table 6.1 Proof of distributive law.
X
Y
Z
Y +Z
X Y
X Z
X ( Y +Z)
XY+XZ
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
1
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
1
1
0
1
1
1
1
1
0
1
1
0
1
1
1
1
1
1
1
1
1
1
