Boolean negation

Question

One of my exam questions reads:

! ( ! ( a != b)  &&  ( b > 7 ) )

The choices:

a) (a != b) || (b < 7)
b) (a != b) || (b <= 7)
c) (a == b) || (b <= 7)
d) (a != b) && (b <= 7)
e) (a == b) && (b > 7)

Initially, I thought it would be D. This is incorrect, and I realize why. I don't understand how the logical negation operator reverses && and greater than/less than. I believe I have narrowed it down to the first two. Is there any instance > would change to <= ?

Answer 1

Is there any instance > would change to <= ?

Answer: every time you negate it.

Consider x > 1. The negation of this is clearly x <= 1. If you simply negate it as x < 1 then neither case covers the x == 1 case.

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That being said, the given boolean ! ( ! ( a != b) && ( b > 7 ) ) can be decomposed as follows:

1. Given:

   ! ( !(a != b) && (b > 7))

2. Negate a != b:

   ! ((a == b) && (b > 7))

3. Distribute the !:

   !(a == b) || !(b > 7)

4. Negate a==b:

   (a != b) || !(b > 7)

5. Negate b>7:

   (a != b) || (b <= 7)

The answer is, therefore, B.

Answer 2

The answer should be B. This is because the negation next to the (a != b) is evaluated first, then you distribute the outside negation to the entire proposition.

Using DeMorgan's Laws, the && will switch to ||. Similarly, != becomes ==, and > becomes <=.

!(!(a != b) && (b > 7))
!((a == b) && (b > 7))
 (a != b) || (b <= 7)