Hello I am trying to simplify this expression (proving consensus expression):
(a + b)(b'+ c)(a + c) = (a + b)(b'+ c)
I was thinking of adding (a+b)(b'+ c)(a + c + b' + b), but I don't know what to do after.
Take a look at this:
(a + b)(b' + c)(a + c)
= (ab' + 0 + ac + bc)(a + c)
= (ab' + ab'c +ac + ac + abc + bc)
= (ab' + ab'c + ac + abc + bc)
= (ab'(1+c) + ac + bc(a + 1))
= (ab' + ac + bc)
= (ab' + c(a+b))
= (ab' + bb' + c (a+b))
= (a+b)(b' + c)
The key step is realising that bb' = 0 so you can safely add that term without affecting the result in the penultimate step.
(A+C')(B'+C')
AB'+AC'+BC'+C'
AB'+(A+B+1)C'
We Know ( 1+anything =1)
So required expression AB'+C' .
Related
I'm working on some logic homework and I can't figure out the next step in reducing the number of literals. Any help would be greatly appreciated.
(A + B + C) (A’B’ + C)
A’B’C + AC + BC + C
C(A’B’ + A + B + C)
C((A + B)’ + A + B + C)
I'm pretty sure I use the associative law next, but I don't understand how the not operator is distributed when rearranging.
From the point where you left:
C((A + B)’ + A + B + C)
C(1 + C) ; X' + X = 1 applied to X = A + B
C(1) ; 1 + <anything> = 1
C ; <anything>1 = <anything>
Hi, I am junior in college and having trouble with my computer architecture classwork. Anyone care to help & tell me if I got them right?
Question1. Convert truth table into bool equation.
Question2. Find miminum SOP(sum of products)
Question3. Use K-map(Karnaugh map) to simplify.
You can simplify the original expression matching the given truth-table just by using Karnaugh maps:
f(x,y,z) = ∑(1,3,4,6,7) = m1 + m3 + m4 + m6 + m7
= ¬x·¬y·z + ¬x·y·z + x·y·z + x·¬y·¬z + x·y·¬z //sum of minterms
f(x,y,z) = ∏(0,2,5) = M0 · M2 · M5
= (x + y + z)·(x + ¬y + z)·(¬x + y + ¬z) //product of maxterms
f(x,y,z) = x·y + ¬x·z + x·¬z //minimal DNF
= (x + z)·(¬x + y + ¬z) //minimal CNF
You would get the same result using the laws of Boolean algebra:
¬x·¬y·z + ¬x·y·z + x·y·z + x·y·¬z + x·¬y·¬z
¬x·(¬y·z + y·z) + x·(y·z + y·¬z + ¬y·¬z) //distributivity
¬x·(z·(¬y + y)) + x·(y·(z + ¬z) + ¬y·¬z)) //distributivity
¬x·(z·( 1 )) + x·(y·( 1 ) + ¬y·¬z)) //complementation
¬x·(z ) + x·(y + ¬y·¬z)) //identity for ·
¬x·(z ) + x·(y + y·¬z + ¬y·¬z)) //absorption
¬x·(z ) + x·(y + ¬z·(y + ¬y)) //distributivity
¬x·(z ) + x·(y + ¬z·( 1 )) //complementation
¬x·(z ) + x·(y + ¬z) //identity for ·
¬x·z + x·y + x·¬z //distributivity
¬x·z + x·y + x·¬z //minimal DNF
¬x·z + x·y + x·¬z
¬x·z + x·(y + ¬z) //distributivity
(¬x + x)·(¬x + (y + ¬z))·(z + x)·(z + (y + ¬z)) //distributivity
( 1 )·(¬x + y + ¬z )·(z + x)·(z + y + ¬z) //complementation
( 1 )·(¬x + y + ¬z )·(z + x)·(y + 1) //complementation
( 1 )·(¬x + y + ¬z )·(z + x)·(1) //annihilator for +
(¬x + y + ¬z )·(z + x) //identity for ·
(¬x + y + ¬z)·(x + z) //minimal CNF
{a(b+c)+a’b}’
using demorgans theorem I got a'+ b'c'a + b' then I factored b' out of b'c'a + b' to get b'(1+c'a) which just turns into b'. plugging it back into the equation I'm left with a'+b'. Is that correct or do I have this all wrong?
{a(b+c)+(a'b)}' = (a (b+c))' . (a'b)'
= (a' + (b+c)') . (a+b')
= (a' + (b'.c')) . (a+b')
= (a.a') + (a'b') + (ab'c') + (b'c')
= 0 + a'b' + b'c'(a+1)
= a'b' + b'c'(1)
= a'b' + b'c'
= b'(a'+c')
Need help here, whats the simplified form of this Boolean expression?
I'm a little confused about this, help me guys!
A'BC + AB'C + A'B'C' + AB'C + ABC
Assuming the ⋅ operator represents binary conjunction, the + binary disjunction and the ' or the ¬ unary negation, you can apply the laws of Boolean algebra:
¬a⋅b⋅c + a⋅¬b⋅c + ¬a⋅¬b⋅¬c + a⋅¬b⋅c + a⋅b⋅c
¬a⋅b⋅c + a⋅¬b⋅c + ¬a⋅¬b⋅¬c + a⋅b⋅c //idempotence of +: a⋅¬b⋅c + a⋅¬b⋅c = a⋅¬b⋅c
b⋅c⋅(¬a + a) + a⋅¬b⋅c + ¬a⋅¬b⋅¬c //distributivity:¬a⋅b⋅c + a⋅b⋅c = b⋅c⋅(¬a + a)
b⋅c⋅(1) + a⋅¬b⋅c + ¬a⋅¬b⋅¬c //complementation: ¬a + a = 1
b⋅c + a⋅¬b⋅c + ¬a⋅¬b⋅¬c //identity for ⋅: b⋅c⋅(1) = b⋅c
b⋅c + a⋅b⋅c + a⋅¬b⋅c + ¬a⋅¬b⋅¬c //absorption: b⋅c = b⋅c + a⋅b⋅c
b⋅c + a⋅c⋅(b + ¬b) + ¬a⋅¬b⋅¬c //distributivity: a⋅b⋅c + a⋅¬b⋅c = a⋅c⋅(b + ¬b)
b⋅c + a⋅c⋅(1) + ¬a⋅¬b⋅¬c //complementation: b + ¬b = 1
b⋅c + a⋅c + ¬a⋅¬b⋅¬c //identity for ⋅: a⋅c⋅(1) = a⋅c
The final line is the original expression's minimal DNF. You can also transform it to it's minimal CNF:
b⋅c + a⋅c + ¬a⋅¬b⋅¬c
(b⋅c + a⋅c) + (¬a⋅¬b⋅¬c)
((b⋅c + a⋅c) + ¬a)⋅((b⋅c + a⋅c) + ¬b)⋅((b⋅c + a⋅c) + ¬c) //distributivity
(b⋅c + a⋅c + ¬a) ⋅ (b⋅c + a⋅c + ¬b) ⋅ (b⋅c + a⋅c + ¬c)
(b⋅c + c + ¬a) ⋅ (b⋅c + a⋅c + ¬b) ⋅ (b⋅c + a⋅c + ¬c) //absorption
(b⋅c + c + ¬a) ⋅ (c + a⋅c + ¬b) ⋅ (b⋅c + a⋅c + ¬c) //absorption
(b⋅c + c + ¬a) ⋅ (c + a⋅c + ¬b) ⋅ (b + a + ¬c) //absorption
(c + ¬a) ⋅ (c + a⋅c + ¬b) ⋅ (b + a + ¬c) //absorption
(c + ¬a) ⋅ (c + ¬b) ⋅ (b + a + ¬c) //absorption
(¬a + c)⋅(¬b + c)⋅(a + b + ¬c)
For this small number of variables, you can also use Karnaugh maps. In the picture (generated using latex) you can see marked out three equivalent expressions – the original one, it's minimal DNF and it's minimal CNF:
f(a,b,c) = ¬a⋅b⋅c + a⋅¬b⋅c + ¬a⋅¬b⋅¬c + a⋅b⋅c
= b⋅c + a⋅c + ¬a⋅¬b⋅¬c
= (¬a + c)⋅(¬b + c)⋅(a + b + ¬c)
I have a boolean simplification problem that's already been solved.. but I'm having a hard time understanding one basic thing about it.. the order in which it was solved.
The problem is simplifying this equation:
Y = ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
The solution is:
Y = ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
= ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + A¬BC + ABC (idempotency for A¬BC)
= ¬A¬C(¬B + B) + A¬B(¬C + C) + AC(¬B + B)
= ¬A¬C + A¬B + AC
The way I solved it is:
Y = ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
= ¬A¬B¬C + ¬AB¬C + ¬A¬B¬C + A¬B¬C + A¬BC + ABC (idempotency for ¬A¬B¬C)
= ¬A¬C(¬B + B) + ¬B¬C(¬A + A) + AC(¬B +B)
= ¬A¬C + ¬B¬C + AC
So how do I know which term to use the law of idempotency on? Thanks.
¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
¬A¬C(¬B + B) + A(¬B¬C + ¬BC + BC)
¬A¬C + A(¬B¬C + ¬BC + BC) <- see truth table below for the simplification of this
¬A¬C + A(¬B + C)
¬A¬C + A¬B + AC
truth table:
B C
0 0 = 1 + 0 + 0 = 1
0 1 = 0 + 1 + 0 = 1
1 0 = 0 + 0 + 0 = 0
1 1 = 0 + 0 + 1 = 1
which is ¬B + C