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6.1 DIGITAL BUILDING BLOCKS 279

Solution

The solution is given in Figure E6.1.4(b).

A

C F

B

(a)

Figure E6.1.4

A

A

B

A ⋅ C

F = A ⋅ B + A ⋅ C + B ⋅ C

A ⋅ B

B ⋅ C

F

C

B

C

(b)

EXAMPLE 6.1.5

Prove the Boolean identityA(B+C)=AB+AC, which is distributive, by comparing the truth
tables of both sides.

Solution

The truth tables with the necessary intermediate variables are as follows:

ABCB+CA(B+C) AB AC AB+AC

000 0 0 00 0

001 1 0 00 0

010 1 0 00 0

011 1 0 00 0

100 0 0 00 0

101 1 1 01 1

110 1 1 10 1

111 1 1 11 1

↑ Same ↑

Note:The direct representation ofAB+ACwould require three gates (two ANDs and one
OR). But two gates (one OR and one AND) can perform the same function, as seen from the
identity.