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.
Related
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
I know how to convert first and second term to the first term of the simplified expression, but I don't know how to convert the rest.
By simplifying, I can get rid of A_Bar in the third term and A in the fifth term and get =B*C_bar
How is it that B*C_bar + the fourth term = becomes XOR(B,C) ?
The two expressions are clearly the same. This can be easily proven by truth tables.
The first one is:
And the second one:
However, this does not fully answer your question.
B*C_bar + the fourth term = becomes XOR(B,C)
This is clearly true if A is true, since per definitionem, B XOR C = B_bar and C OR B and C_bar.
If A is false, these terms are always false and you cannot simplify these two to B XOR C! They are not equal!
Note: Tables generated with http://web.stanford.edu/class/cs103/tools/truth-table-tool/
Note2: ^= OR, ¬ = NOT, ∨ = AND
let play a game.
Let a=not(A), b=not(B) and c=not(C) and *=xor
Y = ab + (B*C)
Y = ab + Bc + bC
Y = ab(1) + Bc(1) + bC(1)
Y = ab(c+C) + Bc(a+A) + bC(a+A)
Y = abc + abC + Bca + BcA + bCa + bCA
Y = abc + abC + aBc + ABc + abC + AbC
Y = abc + abC + aBc + ABc + AbC
That is the first equ.
Does anybody know how to find the dual of the following (Boolean Algebra)?
p*(q+r')
Where * stands for AND, + stands for OR, and ' stands for NOT.
I have the following, but it is incorrect:
p + (qr')
not a and not b => not (a or b)
Here a = p', b = (q+r')' = (q'r). Hence p*(q+r') = (p'+q'r)'
how can i specify a mathematical expression using Z notation ?
I think free types is appropriate for this situation because expressions has a recursive nature. please consider that we can have parenthesis and variables in our expression. and only ( + , - , / , * ) is allowed. for example :
A + 2 * ( 3 - B ) / 4
please help me ...
You need to use the axiomatic definition: the definition of an object is constrained by conditions.
There is a schema specified in zet for this.
Which is
| Declaration
------------------------------
| Predicate
|
|
A recursive example given:
[USERNAME] - An already defined type.
Given a username and a sequence of usernames(N1 --> USERNAME) return the number that the given username appears into the sequence.
|-occurs- USERNAME X seq USERNAME → N //here you define the input and what is returned.
---------------------------------------
|∀ u: USERNAME, s: seq USERNAME then
|s = < > => occurs(u,s) = 0
|s ≠ < > and head(s) = u => occurs(u,s) = 1+occurs(u,tail(s))
|s ≠ < > and head(s) ≠ u => occurs(u,s) = occurs(u,tail(s))
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'