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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
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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
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I need to figure out what number base makes the expression 32 + 12 = 28 true. It looks like the second number is signed, I've tried converting to decimal and using 10's complement but haven't gotten an answer that makes sense.
If this is a positional number system, then you can replace the numerals with an expression showing how their digits multiply by the meaning of their columns.
That is,
"32 + 12 = 28"
≣ (3*R^1 + 2*R^0) + (1*R^1 + 2*R^0) = (2*R^1 + 8*R^0)
≣ (3*R + 2*1) + (1*R + 2*1) = (2*R + 8*1)
≣ (3*R + 2) + (1*R + 2) = (2*R + 8)
≣ 4*R + 4 = 2*R + 8
≣ 2*R = 4
≣ R = 2
For some radix R.
You can solve this equation to get the value of R. If you attempt to solve it, you'll see that you get R = 2, which clearly can't be the case, since you have a digit 3 (a base-2 system would only have ... 2 digits. 0, and 1).
Thus, there is no radix R that would allows "32 + 12 = 28" to encode a valid equation.
On the face of it, it looks impossible. Adding positive integers in any base should result in a larger integer; if the second integer is negative, you end up with 20 = 28 which is clearly impossible. Considered modulo arithmetic? 32 mod 4 is 0, 12 mod 4 is 0, 28 mod 4 is 0, 0 + 0 = 0.
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 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.
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