Boolean algebra

hi, i was going through boolean algebra and i could not figure it out. let me know if you have any idea.

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

Well presumeably ' means NOT.

Thus:

F = (A * NOT(B + NOT C)) + ((B + D) * NOT A)

So...

Code:

---

; left of second "+"
mov eax, C
not eax
add eax, B
not eax
mul A
mov ecx, eax ; save result

; right
mov eax, B
add eax, D
mov edx, A
not edx
mul edx
add eax, ecx

; EAX == F (the answer)

---

I could be wrong of course. :D

I mean to say how do you solve this function using boolean algebra?
F= A(B+C')' + (B+D)A'

This is not the math forum.

F = A(B+C')' + (B+D)A'
F = A.B + A.C' + A'.B + A'.D
F = B(A+A') + A.C' + A'.D
F = B + A.C' + A'.D [Because A+A'=1]

Do you mean "How do you simplify the expression" from it's minterms? If so, then google for Karnaugh maps. I don't think we have a thread on discrete design . .. . .. .yet.

(BTW Wossy: Karnaugh maps can be very useful for optimising long sequences of boolean maths in any language)