what are the steps to simplifying this (a+b)(a+!b)=a
(a + b).(a + !b)
= a.(b + !b) ; distributivity [1]
= a.1 ; complements [1]
= a
See Wikipedia page on Boolean algebra
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I have to draw the simplified logic circuit of this given [(A’B’)’ + (A’+ B’)’]’ and also get the simplified boolean expression
From DeMorgan Theory
((A'B')' + (A'+ B')')'=(A+B + AB)'
(A+B + AB)' = (A+B)'. (AB)'
(A+B)'. (AB)' = (A'.B') .(A'+B')
assume
X=A' ,Y=B'
we can conclude that
(XY)(X+Y) is (XY)
as both (XY) and (X+Y) has to be 1 to produce 1 at output as seen from truth table of (XY)(X+Y) is (XY) is identical to XY
so as final result
((A'B')' + (A'+ B')')'=(A+B + AB)'
(A+B + AB)' = (A+B)'. (AB)'
(A+B)'. (AB)' = (A'.B') .(A'+B') = A'B'
before simplifying
after simplifying
comparing outputs to make sure
(A'B')' = A + B
Using DeMorgan's theorem (AB)' = A' + B'
(A'+B')' = AB
Again, using DeMorgan's theorem (A+B)' = A'B'
Therefore, now we have the expression:
(A+B + AB)'
Taking A+B as X and AB as Y
(X+Y)' = X'Y'
= (A+B)'·(AB)'
Now, creating a logic circuit is fairly simple for this expression, inputs A and B are fed to a NOR gate and NAND gate simultaneously whose outputs act as input to an AND gate
This is the expression I need to simplify:
[{(AB)'*(BC)'} + (A'C)']'
Right now I have my answer as (AB + BC).(A' + C)
I don't think my answer is right but I'm not sure. IF someone could show steps on how to simplify that would be really appreciated, thanks!
((AB)'(BC)' + (A'C)')' = ((AB)'(BC)')'(A'C)'' ; DeMorgan
= ((AB)'' + (BC)'')(A'C) ; DeMorgan + double negation
= (AB + BC)(A'C) ; double negation
= ABA'C + BCA'C ; distribution
= BCA' ; AA'=0
I would like help simplifying this boolean algebra expression:
B*C + ~A*~B + ~A*~C => A*B*C + ~A
I need to know the steps of how to simplify it to the ABC + ~A
'*' indicates "AND"
'+' indicates "OR"
"~A" indicates "A NOT"
Any help would be appreciated!
Thank you!
For a better view, i'll skip * for conjunction, and use ' for negation.
First you shall expand the 2 term disjunctions: Expand B*C , A'*B' and A'*C'
1) (A + A')BC + A'B'(C + C') + A'(B + B')C'
now distribute the parentheses.
2) ABC + A'BC + A'B'C + A'B'C' + A'BC' + A'B'C'
the fourth term and the last term are the same, A'B'C', so ignore one of them since p + p = p or you can expand the situation for your needs (might be needed for some situations) as in p+p+p+p+....+p = p
3) So now, lets try to search for common terms. See the 2nd term and 5th term, A'BC and A'BC'. Take common parenthesis, A'B(C+C') => A'B.
Do the same for 3rd term and the 4th term, A'B'C and A'B'C'. A'B'(C+C') => A'B' since X+X' = 1.
now we have:
ABC + A'B + A'B'
4) take common parenthesis again, 2nd and 3rd term: A'(B+B')
There you have ABC + A'
BC + A'B' + A'C' => ABC + A'
I need help simplifying the following Boolean expressions using DeMorgan’s law:
a) [ (AB)' + (CD)' ]'
and
b) [(X+Y)' + (X+Y') ]'
Please show some steps so I can do the other ones myself
a)
First step is the outermost negation: distribute it.
((AB)')'*((CD)')'
You see we have double negations which means the expression itself. (p')' = p
therefore
ABCD
[ (AB)' + (CD)' ]' --> ABCD
b)
Distribute the outermost negation:
((X+Y)')'(X+Y')'
get rid of the double negation:
(X+Y)(X+Y')'
again, distribute the negation (the one at the outer part of the expression):
(X+Y)(X'Y)
When you distribute (X+Y), we get
XX'Y + YX'Y
Since there is XX' in the first part of disjunction, the expression XX'Y equals to 0 (False).
Multiple instances of the same thing in an expression is the same thing itself. ppp = p.
Therefore:
0 + YX' --> YX'
[ (X+Y)' + (X+Y') ]' --> YX'
Im sorry for non-formal language:) hope it helps.
Steps are included:
a: [ (AB)' + (CD)' ]' = (AB)'' * (CD)'' = (AB) * (CD) = ABCD
b: [ (X+Y)' + (X+Y') ]' = (X+Y)'' * (X+Y')' = (X+Y) * (X'*Y) .. Simplifying this further relies on the distributive property.
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Prove (A ⊕ B) ⊕ C = A ⊕ (B ⊕ C) using boolean algebra.
I made the truth tables and found the sum of products, but couldnt figure how to show their equal.
I then tried doing
(a xor b) xor c (a' - is NOT(A)/inverse)
(a'b + ab') ⊕ C
c' (a'b + ab') + c[(a'b + ab')']
Couldn't go from there,
A^B = (AB'+A'B)
(AB)' = (A'+B')
(A^B)^C
= (AB'+A'B)C' + (AB'+A'B)'C
= (AB'C'+A'BC')+((AB')'(A'B)')C
= (AB'C'+A'BC')+(A'+B)(A+B')C
= (AB'C'+A'BC')+(A'(A+B')+B(A+B'))C
= (AB'C'+A'BC')+(A'B' + AB)C
= (AB'C'+A'BC'+A'B'C + ABC)
= A(B'C'+BC)+A'(BC'+B'C)
= A(B'C'+BC)+A'(B^C) (1)
(B^C)'
=(BC'+B'C)'
= (BC')'(B'C)'
= (B'+C)(B+C')
= (B'C'+BC) (2)
From (2), the (1) = A(BC'+B'C)' + A'(B^C) = A(B^C)' + A'(B^C) = A^(B^C) #
Firstly define XOR and XNOR:
A^B = AB' + A'B ... (1)
(A^B)' = AB + A'B' ... (2)
Now expand (A^B)C using (1) and (2):
(A^B)C = (A^B)C' + (A^B)'C
= (AB' + A'B)C' + (AB + A'B')C
= AB'C' + A'BC' + ABC + A'B'C
Collect terms and simplify:
= A(B'C' + BC) + A'(BC' + B'C)
= A(B^C)' + A'(B^C)
= A^(B^C)
QED