My Math Forum Boolean algebra simplification using boolean rules

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 September 8th, 2015, 02:17 PM #1 Newbie   Joined: Sep 2015 From: Uk Posts: 1 Thanks: 0 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
 September 8th, 2015, 04:02 PM #2 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 I suggest you start by using Venn diagram to see what is going on. Then work from the rules.
 September 26th, 2016, 09:11 AM #3 Newbie   Joined: Sep 2016 From: Manila Philippines Posts: 2 Thanks: 0 (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

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