Anyone knows Reduce Boolean Expression

Question

Anyone can help me reduce this to 4 literals?

F = ( A + C + D) ( A + C + D') (A + C' + D) ( A + B')

i tested in logic friday the answer was F = C D B' + A.

Answer 1

After finding the mean term you can use Quine McCluskey Technique to solve the experssion. The result will same as K-Map. But it is fast technique for many veriable boolean expression.

Quine McCluskey online solver

Answer 2

Assuming the ⋅ operator represents binary conjunction, the + binary disjunction and the ' or the ¬ unary negation, I would apply the laws of Boolean algebra this way:

(a + c + d)⋅(a + c + ¬d)⋅(a + ¬c + d)⋅(a + ¬b)
((a + d) + (c⋅¬c))⋅(a + c + ¬d)⋅(a + ¬b)        //distributivity
((a + d) + (0))⋅(a + c + ¬d)⋅(a + ¬b)       //complementation: c⋅¬c = 0
(a + d)⋅(a + c + ¬d)⋅(a + ¬b)           //identity for +: (a + d) + (0) = (a + d)
(a) + (d⋅(c + ¬d)⋅¬b)                 //distributivity
(a) + ((d⋅c + d⋅¬d))⋅¬b)              //distributivity: d⋅(c + ¬d) = (d⋅c + d⋅¬d)
(a) + ((d⋅c + 0))⋅¬b)                 //complementation: d⋅¬d = 0
(a) + (d⋅c⋅¬b)                        //identity for +: (d⋅c + 0) = d⋅c

a + ¬b⋅c⋅d

The final line is the minimal DNF. You can also transform it to it's minimal CNF this way:

(a) + (¬b⋅c⋅d)
(a + ¬b)⋅(a + c)⋅(a + d)              //distributivity

For this small number of variables, you could have used Karnaugh maps to quickly find the minimal forms or to control your result. In the following picture (generated using latex) is the original expression next to it's minimal DNF and minimal CNF.