Hello, chillerbros17!

The exclusive-or operation is defined by the following truth table: Code:

p q | p __v__ q
------+-------
T T | F
T F | T
F T | T
F F | F

Exclusive-or means "p or q *but not both*".

a. Express p __v__ q in terms of p, q, v, ^, ~ I would write: .(p ^ ~q) v (~p ^ q)

b. Express p v q in terms of p, q, ^, ~, __v__ Tricky . . . the "or" is not in the list.

So I can't write: .p v q

But I can use DeMorgan's Law to write: .~(~p ^ ~q)

. . (This works: It says "It is not true that both are false".)

c. Show that one of the distributive laws no longer holds when v is replaced by __v__. Compare: . p __v__ (q ^ r) .and ,(p __v__ q) ^ (p __v__ r) Code:

p __v__ (q ^ r) (p __v__ Q) ^ (P __v__ R)
------------ -----------------
T F T T T T F T F T F T
T T T F F T F T F T T F
T T F F T T T F F T F T
T T F F F T T F T T T F
F T T T T F T T T F T T
F F T F F F T T F F F F
F F F F T F F F F F T T
F F F F F F F F F F F F
↑__ not equal __↑