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

 

I'm not sure what you mean by "take it further". Could you elaborate?

 

I'm not sure what you mean by "take it further". Could you elaborate?

Sorry, I should have mentioned I need (A+B)(A+C)(A+D) = A+BCD

 

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

 

(a + b)(a + c)(a + d)
(a + bc)(a + d) [Distributive]
(a + bcd) [Distributive]