solve X' ((Y.Z)' + Y'
Answers
Explanation:
2-1: Demonstrate by means of truthtables
Points to remember:
1. truth tables ARE proof
2. proof by truthtable is time intensive -- it is better to simplify
first.
3. group, then analyze (make it as mechanical as possible)
(a) Demorgan's Theorem for three variables: (XYZ)' = X' + Y' + Z'
X Y Z XYZ (XYZ)' X'+Y'+Z'
0 0 0 0 1 1
0 0 1 0 1 1
0 1 0 0 1 1
0 1 1 0 1 1
1 0 0 0 1 1
1 0 1 0 1 1
1 1 0 0 1 1
1 1 1 1 0 0
Observe that the general form (the negation of some expression
ABCDEF...XYZ) will follow the same pattern. Why?
(b) X + YZ = (X+Y)(X+Z)
(algebraic solution: XX+XY+XZ+YZ = X(1+Y+Z)+YZ = X+YZ)
(speed tip: note that in + sets you should scan & fill in the leftmost
variable first, marking as true, then the rightmost (alternating), and
so on -- e.g., in X+YZ below you can fill in the bottom four 1's
automatically)