Boolean Algebra (Boolean Expression)

Running head: BOOLEAN ALGEBRA
BOOLEAN ALGEBRA
Name of the Student
Name of the University
Author Note

BOOLEAN ALGEBRA1
Solution 1:
AB(A+B)(C+C)
= AB(A+B)C
= AB(A+A’B)C
=(ABA+ABA’)C
= (AB+0B)C
= ABC
Solution 2:
(p,*r)+q
Solution 3:
[(p^q’)v(rvq)]^s

BOOLEAN ALGEBRA2
Solution 4:
P Q R P’ (Q v R) P’ ^ (Q v R)
0 0 0 1 0 0
0 0 1 1 1 1
0 1 0 1 1 1
0 1 1 1 1 1
1 0 0 0 0 0
1 0 1 0 1 0
1 1 0 0 1 0
1 1 1 0 1 0
Solution 5:
Simplifying the first expression, we get,
(A+C) (AD + AD’) +AC +C
=(A+C)A(D+D’)+AC+C
=(A+C)A+AC+C
=(AA+AC)+AC+C
=(A+AC)+AC+C