DHARM
N-COM\APPENDIX.PM5BOOLEAN ALGEBRA 365
= (x + y′) (x′ + 0) (0 + y) P1
= (x + y′) (x′ + y y′) (x x′ + y) P2
= (x + y′) (x′ + y) (x′ + y′) (x + y) (x′ + y) P4
After simplifying we obtain,
= (x + y) (x + y′) (x′ + y) (x′ + y′)
To obtain the PDNF, we proceed like as,
= (y′ + x) (x′ y)
= y′ x′ y + x x′ y P4
= x′ y′ y + x x′ y P3
= x′. 0 + 0. y P2
= 0 + 0 T3
= 0P 1
(2) x → (x. (y → x)) = x → (x. (y′ + x)) Equiv. (i)
= x → (x. y′ + x. x) P4
= x → (x. y′ + x) T1
= x′ + (x. y′ + x) Equiv. (i)
= x′ + x + x y′ P3
= 1 + x y′ P2
= 1 T2
So a PCNF expression is, (x + x′) (y + y′) P2
To obtain PDNF we proceed as,
= (x + x′) (y + y′) P2
= x y + x′ y + x y′ + x′ y′ P4
= x′ y′ + x′ y + x y′ + x y P3
= x′ y′(z + z′) + x′ y (z + z′) + x y′(z + z′) + x y (z + z′)
= x′ y′ z + x′ y′ z′ + x′ y z + x′ y z′ + x y′ z + x y′ z′ + x y z + x y z′ P4
PDNF
(3) exercise to readers.
(4) Let f(x, y, z) = (x′ → z) (y ↔ x)
= ((x′)′ + z) (y → x) (x → y) Equiv. (i) & (ii)
= (x + z) (y → x) (x → y) T3
= (x + z) (y′ + x) (x′ + y) Equiv. (i)
= (x + z) (x + y′) (x′ + y) P3
= (x + 0 + z) (x + y′ + 0) (x′ + y + 0) P1
= (x + y y′ + z) (x + y′ + z z′) (x′ + y + z z′) P2
= (x + y + z) (x + y′ + z) (x + y′ + z) (x + y′ + z′) (x′ + y + z) (x′ + y + z′) P4
= (x + y + z) (x + y′ + z) (x + y′ + z′) (x′ + y + z) (x′ + y + z′) Simp.
PCNF
Find the PDNF,
= (x′ → z) (y ↔ x)
= (x + z) (x + y′) (x′ + y)