Simplifying equations using boolean algebra


by kirti.1127
Tags: algebra, boolean, equations, simplifying
kirti.1127
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#1
Nov19-07, 03:18 AM
P: 10
[b]1. Simplify the following equations using boolean algebra

[b]2. a) abc + ab'c'+ ab'c

b) (abc)'+(a+c)'+b'c'


[b]3. Please help me to solve the above equations
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mjsd
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#2
Nov19-07, 04:30 AM
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pls first clarify your notations.
I suspect abc means a AND b AND c?
a+b means a OR b?
a' means NOT a?
right?

now: what standard identities do u know?
kirti.1127
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#3
Nov19-07, 10:06 PM
P: 10
ya. I could not use the complement notation while posting the query. I know all the standard notations. In second problem first i used sop and pos rules but got stuck up on next step. So please if u can help me in that.

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#4
Nov21-07, 03:26 AM
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Simplifying equations using boolean algebra


Quote Quote by kirti.1127 View Post
ya. In second problem first i used sop and pos rules but got stuck up on next step.
sop? pos?
what's that? anyway,

have u tried playing around using the standard axioms:
eg. associativity, commutativity, distributivity, De Morgan's Law, idempotence...etc.

for example:
a.a = a, a+a=a, (a.b).c = a.(b.c), a+b =b+a, a.(b+c)=a.b+a.c, a+(b.c)=(a+b).(a+c)
a+a'=1, a.a'=0
etc.
kirti.1127
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#5
Nov21-07, 10:19 PM
P: 10
sop-sum of product rule
and pos means product of sum rule.

Ya i tried.
mjsd
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#6
Nov22-07, 05:32 AM
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try reversing the distributive law to gather "common factor"
repeat use of axioms etc. it is a bit of a trial and error process, unless you can see something in advance (which comes with experience only). But you can always check, at each step, that you have not make an error by checking the truth table for both the original and derived expression.
another hint, sometimes it may even be useful to "add terms" into your expression


of course, sometimes it may be difficult to tell whether you have reduced your expression as simple as possible.
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#7
Nov22-07, 05:37 AM
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another hint:
x.y + y = y
since
x.y+y = (x+1).y = (1).y = y
kirti.1127
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#8
Nov22-07, 10:17 PM
P: 10
I tried solving the first pb:
solution:

abc+ab'c'+ab'c
=abc+a(b'+c')+ab'c (Product of Sum rule)
=abc+ab'+ac'+ab'c
=abc+ac'+ab'(1+c)
=abc+ac'+ab' (1+c=1)
=abc+a(c'+b')
=abc+ab'c' (Sum of Product rule)

Just check if the solution is correct
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#9
Nov23-07, 12:13 AM
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Quote Quote by kirti.1127 View Post
abc+ab'c'+ab'c
=abc+a(b'+c')+ab'c (Product of Sum rule)
?? b'c' => b'+c' ?

is it b'c' or (b.c)' ?

you can always check answer by simply writing out the truth table for the original expression and then compare with the one for the new expression.
kirti.1127
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#10
Nov23-07, 12:51 AM
P: 10
Oh ya u r correct.
that's y i told u to check.


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