Boolean Algebra Help

Thread Starter

Thirsty

Joined Aug 9, 2006
5
Hi, can anyone help me take this further please....

(a+b)(a+c)(a+d)

(aa+ac+ba+bc)(a+d)

aaa+aad+aca+acd+baa+bad+bca+bcd
 

Dave

Joined Nov 17, 2003
6,970
Remember:

A.A = A

So you should be able to simplify your final line to:

a+ad+ac+acd+ba+bad+bca+bcd

a(d+c+cd+b+bd+bc) + bcd

De Morgans:

C.D = C' + D'

a(d+c+c'+d'+b+b'+d'+b'+c') + bcd

And:

B' + B' = B'

So simplifying:

a(d+c+c'+d'+b+b') + bcd

And:

C + C' = 1

Giving:

a + bcd

Dave
 
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