Integer Variables - Which Three Logical Expressions Are Equivalent To Each Other

Assuming that x, y, and z are integer variables, which of the following three logical expressions are equivalent to each other, that is, have equal values for all possible values of x, y, and z?
 
(x == y && x != z) || (x != y && x == z)
(x == y || x == z) && (x != y || x != z)
(x == y) != (x == z)
 
None of the three
 
A. I and II only 
B. II and III only 
C. I and III only
  D. I, II, and III

I selected B, but got it wrong. I really think I need understanding boolean logic. The correct answer says something else but I don't get the logic. Here is the correct answer:

Answer Key : The following model answer has been provided to you by the grader. Carefully compare your answer with the one provided here.

Expression III is the key to the answer: all three expressions state the fact that exactly one out of two equalities, x == y or x == z, is true. Expression I states that either the first and not the second or the second and not the first is true. Expression II states that one of the two is true and one of the two is false. Expression III simply states that they have different values. All three boil down to the same thing. The answer is E.

In exercise 4, I get the same problem:

The expression !((x <= y) && (y > 5)) is equivalent to which of the following?

A. (x <= y) && (y > 5)
B. (x <= y) || (y > 5)
C. (x >= y) || (y < 5)
D. (x > y) || (y <= 5)
E. (x > y) && (y <= 5)

Exercise 4
ABCDE
Incorrect
Score: 0 / 1
Submitted: 2/10/2014 8:21pm
Your answer is incorrect.
Answer Key

The following model answer has been provided to you by the grader. Carefully compare your answer with the one provided here. The given expression is pretty long, so if you try to plug in specific numbers you may lose a lot of time. Use De Morgan's Laws instead:

!((x <= y) && (y > 5))
 !(x <= y) || !(y > 5)

When ! is distributed,
&& changes into ||, and vice-versa

(x > y) || (y <= 5)