ANSWERS TO BOOLEAN LOGIC EXERCISES
1. EQ(x, y) = xy + x'y'
x y EQ(x,y)
0 0 1
0 1 0
1 0 0
1 1 1
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2. (a) .jpg)
(b) 
Note: these two Boolean circuits are equivalent.
3. x y z f
0 0 0 1 f = x'y'z' + xyz
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
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4. x y z g
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0
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The circuit above is created using the sum-of-products method.
Observe that a simpler circuit can be obtained by adding a NOT
gate to the circuit in the previous question since g = f'.
5. x y z w f x y z w f
0 0 0 0 0 1 0 0 0 0
0 0 0 1 0 1 0 0 1 1
0 0 1 0 0 1 0 1 0 1
0 0 1 1 1 1 0 1 1 0
0 1 0 0 0 1 1 0 0 1
0 1 0 1 1 1 1 0 1 0
0 1 1 0 1 1 1 1 0 0
0 1 1 1 0 1 1 1 1 0
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f = x'y'zw + x'yz'w + x'yzw' + xy'z'w + xy'zw' + xyz'w'
6. (a) 
(b) 
(c) 
7. (a) x2'x1'x0 + x2'x1 x0' + x2 x1'x0'
(b) x2'x1'x0' + x2'x1'x0 + x2'x1 x0'
(b) x2 x1'x0' + x2 x1'x0 + x2 x1 x0'
8. The decoder implements all possible N-way "products" among its N inputs
and their negations. Since we can implement any Boolean function using
sum-of-products, the only other component we need is an (N-way) OR gate.
By passing the appropriate decoder outputs through an OR gate, we get the
appropriate "sum" term.
9. Draw the truth table and use sum-of-products.
A B S | C D C = A'BS' + AB'S + ABS' + ABS
0 0 0 | 0 0 D = A'BS + AB'S' + ABS' + ABS
0 0 1 | 0 0
0 1 0 | 1 0
0 1 1 | 0 1
1 0 0 | 0 1
1 0 1 | 1 0
1 1 0 | 1 1
1 1 1 | 1 1
10. 