Reputation: 491
For my class I have to study some Boolean algebra. Now I'm having some difficulties with simplifying expression.
For example I get:
A.B.C + NOT(A) + NOT(B) + NOT(C)
I tried checking wolfram alpha but there's not simplification showing up there. Can you tell me how to simplify this expression?
Thanks in advance
Upvotes: 1
Views: 10947
Reputation: 617
Use http://en.wikipedia.org/wiki/De_Morgan%27s_laws
A AND B = NOT(NOT(A) OR NOT(B))
A OR B = NOT(NOT(A) AND NOT(B))
and the normal methods for distribution, commutation, etc. See http://en.wikipedia.org/wiki/Boolean_algebra#Laws
Note that in the linked texts above the symbols are written differently.
Upvotes: 0
Reputation: 111491
Given Boolean expression:
abc + a' + b' + c'
Apply double negation:
(abc + a' + b' + c')''
Apply De Morgan's Law for a disjunction:
((abc)'a''b''c'')'
Reduce double negations:
((abc)'abc)'
AND of x and x' is 0:
(0)'
Negation of 0 is 1:
1
Given Boolean expression:
a.b.c + NOT(a) + NOT(b) + NOT(c)
Apply double negation:
NOT(NOT(a.b.c + NOT(a) + NOT(b) + NOT(c)))
Apply De Morgan's Law for a disjunction:
NOT(NOT(a.b.c).NOT(NOT(a)).NOT(NOT(b)).NOT(NOT(c))))
Reduce double negations:
NOT(NOT(a.b.c).a.b.c)
AND of x and x' is 0:
NOT(0)
Negation of 0 is 1:
1
Upvotes: 2
Reputation: 4508
Wolfram Alpha wasn't giving a simplification because it didn't understand your notation. Using (A and B and C) or NOT(A) or NOT(B) or NOT(C)
shows that it simplifies to true.
Or you can just look at it: if any are false, the NOT
will makke everything true, and if they're all true, then so is the first clause.
Upvotes: 1
Reputation: 212929
Truth table:
A B C X
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
So the simplification is just:
X = 1
Upvotes: 3