Sederhanakan Ekspresi Logika Dibawah Ini Dengan Aljabar Boolean

1.	AB’ + BC + C’A	= AB’ + BC + AC’ = A(B’ + C’) + BC = A.1 = A	(T1 b) (T3a) (T2 b)
2.	A’ (BC + AB + BA’)	= A’.(B (C +A + A’)) = A’.(B(C+1)) = A’.(B.1) = A’.B	(T1 a) (T8b) (T7C) (T7 b)
3.	ABC + AB + A	= ABC + AB + A.1 = A.(BC + B + 1) = A.(B.(C+1)+1) = A.(B.1+1) = A.(B +1) = A.(1) =A	(T7 b) (T3a) (T7 b) (T7 C) (T7b) (T7C) (T7b)
4.	(A’ + AB).(A’B)	=A’.(A’B) +B(AA’) =A’.(A’B) +B(0) =A’.(A’B) +0 =A’.(A’B) = A’B	(T3a) (T8b) (T7d)
5.	BC + AC + ABCD + ADC + A’	= BC.(1+AD) + AC.(1+D)+A’ = BC.(1) + AC.(1) + A’ = BC + AC + A’	(T7b) (T3a)  (T7C) (T7b)

BUATLAH TEBEL KEBENARAN DARI PERSAMAAN LOGIKA DIBAWAH:

a.         X.Y + X’.Y + X’.Y’ = X’Y

X	Y	X’	Y’	X.Y	X’.Y	X’.Y’	X.Y + X’Y	(X.Y + X’Y) + X’.Y’	X’+Y
1	1	0	0	1	0	0	1	1	1
1	0	0	1	0	0	0	0	0	0
0	1	1	0	0	1	0	1	1	1
0	0	1	1	0	0	1	0	1	1

b.        A.B.C + A.C + B.C = (A+B).C

A	B	C	A+B	A.B	A.C	B.C	(A.B).C	(A.B.C)+(A.C)	(A.B.C)+(A.C)+(B.C)	(A+B).C
1	1	1	1	1	1	1	1	1	1	1
1	1	0	1	1	0	0	0	0	0	0
1	0	1	1	0	1	0	0	1	1	1
0	0	0	0	0	0	0	0	0	0	0
0	1	1	1	0	0	1	0	0	1	1
0	1	0	1	0	0	0	0	0	0	0

c.         (X’.Y + Y’.X) + X.Y = (X+Y)

X	Y	X’	Y’	X.Y	X’.Y	X.Y’	(X’.Y)+( X.Y’)	(X’.Y+X.Y’)+ (X.Y)	X+Y
1	1	0	0	1	0	0	0	1	1
1	0	0	1	0	0	1	1	1	1
0	1	1	0	0	1	0	1	1	1
0	0	1	1	0	0	0	0	0	0