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September 8th, 2015, 02:17 PM   #1
Joined: Sep 2015
From: Uk

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Boolean algebra simplification using boolean rules

Hey guys,
First off, sorry for the overly easy question.
I need to prove this boolean expression using the Boolean logic rules.
(Commutative, Associative, Distributive, Identity, Redundance, De Morgan...)

(a or b) and (c or (not b)) = (a and (not b)) or (b and c)
(a + b)(c + ¬b) = a(¬b) + bc

I really can't figure a way around it. It seams pretty simple so i figured i'm probably missing something quite obvious...
I keep ending up with:
ca + bc + a(¬b) = a(¬b) + bc
but i can't figure out how to get rid of ca.

The other partial solution i had come across was:
(a + b)(c + (¬b)) = ¬b(a + b) + b(c + ¬b)

I spent quite a lot of time on it and eventually gave up.
Please tell me what i'm doing wrong.
Thanks in advance
JustForFun2000 is offline  
September 8th, 2015, 04:02 PM   #2
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I suggest you start by using Venn diagram to see what is going on. Then work from the rules.
mathman is offline  
September 26th, 2016, 09:11 AM   #3
Joined: Sep 2016
From: Manila Philippines

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(a + b)(c + ¬b) = a(¬b) + bc
(A+B)(C+B') = AB' + BC

after FOIL
AC + AB' + BC + BB' = AB' + BC
AC + AB' + BC = AB' + BC

expand AC
ABC + AB'C + AB' + BC = AB' + BC

extract AB' and BC
AB'(1+C) + BC(1+A) = AB' + BC
AB' + BC = AB' + BC
Darwin1521 is offline  

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