Boolean Expression
Chapter 2
Examples of Solved Problems
This section presents some typical problems that the student may encounter, and shows how such problems can be solved. In addition to the identities given in Section 2.5, these examples also use an identity known as consensus, defined below.
17a.
17b.
x·y+y·z+x·z =x·y+x·z
(x + y) · (y + z) · (x + z) = (x + y) · (x + z)
Consensus
Example 2.1
Problem: Determine if the following equation is valid x1 x3 + x2 x3 + x1 x2 = x1 x2 + x1 x3 + x2 x3
Solution: The equation is valid if the expressions on the left- and right-hand sides represent the same function. To perform the comparison, we could construct a truth table for each side and see if the truth tables are the same. An algebraic approach is to derive a canonical sum-of-product form for each expression.
Using the fact that x + x = 1 (Theorem 8b), we can manipulate the left-hand side as follows:
LHS = x1 x3 + x2 x3 + x1 x2
= x1 (x2 + x2 )x3 + (x1 + x1 )x2 x3 + x1 x2 (x3 + x3 )
= x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3
These product terms represent the minterms 2, 0, 7, 3, 5, and 4, respectively.
For the right-hand side we have
RHS = x1 x2 + x1 x3 + x2 x3
= x1 x2 (x3 + x3 ) + x1 (x2 + x2 )x3 + (x1 + x1 )x2 x3
= x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3
These product terms represent the minterms 3, 2, 7, 5, 4, and 0, respectively. Since both expressions specify the same minterms, they represent the same function; therefore, the equation is valid.
Another way of representing this function is by m(0, 2, 3, 4, 5, 7).
1
Example 2.2
Problem: Design the minimum-cost product-of-sums expression for the function f (x1 , x2 , x3 , x4 ) = m(0, 2, 4, 5, 6, 7, 8, 10, 12, 14, 15).
Solution: The function is defined in terms of its minterms. To find a POS expression we should start with the definition in terms of maxterms, which is f = ΠM (1, 3, 9, 11, 13). Thus, f = M1 · M3 · M9 · M11 · M13
= (x1 + x2 + x3 +
Examples of Solved Problems
This section presents some typical problems that the student may encounter, and shows how such problems can be solved. In addition to the identities given in Section 2.5, these examples also use an identity known as consensus, defined below.
17a.
17b.
x·y+y·z+x·z =x·y+x·z
(x + y) · (y + z) · (x + z) = (x + y) · (x + z)
Consensus
Example 2.1
Problem: Determine if the following equation is valid x1 x3 + x2 x3 + x1 x2 = x1 x2 + x1 x3 + x2 x3
Solution: The equation is valid if the expressions on the left- and right-hand sides represent the same function. To perform the comparison, we could construct a truth table for each side and see if the truth tables are the same. An algebraic approach is to derive a canonical sum-of-product form for each expression.
Using the fact that x + x = 1 (Theorem 8b), we can manipulate the left-hand side as follows:
LHS = x1 x3 + x2 x3 + x1 x2
= x1 (x2 + x2 )x3 + (x1 + x1 )x2 x3 + x1 x2 (x3 + x3 )
= x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3
These product terms represent the minterms 2, 0, 7, 3, 5, and 4, respectively.
For the right-hand side we have
RHS = x1 x2 + x1 x3 + x2 x3
= x1 x2 (x3 + x3 ) + x1 (x2 + x2 )x3 + (x1 + x1 )x2 x3
= x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3
These product terms represent the minterms 3, 2, 7, 5, 4, and 0, respectively. Since both expressions specify the same minterms, they represent the same function; therefore, the equation is valid.
Another way of representing this function is by m(0, 2, 3, 4, 5, 7).
1
Example 2.2
Problem: Design the minimum-cost product-of-sums expression for the function f (x1 , x2 , x3 , x4 ) = m(0, 2, 4, 5, 6, 7, 8, 10, 12, 14, 15).
Solution: The function is defined in terms of its minterms. To find a POS expression we should start with the definition in terms of maxterms, which is f = ΠM (1, 3, 9, 11, 13). Thus, f = M1 · M3 · M9 · M11 · M13
= (x1 + x2 + x3 +