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 alot!
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 alot!