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In which situation would the expression (!x || y) || !(!x || y) evaluate to true?

1) x and y both have the initial value true
2) x and y both have the initial value false
3) x has the value true and y has the value false
4) The expression would never evaluate to true

User Teqnology
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1 Answer

2 votes

Final answer:

The expression (!x || y) || !(!x || y) evaluates to true in three different situations: when x and y are both true, when x and y are both false, and when x is true and y is false.

Step-by-step explanation:

To determine when the expression (!x || y) || !(!x || y) evaluates to true, we can consider the different cases:

  1. If both x and y have the initial value true, the expression will evaluate to true. This is because when x is true, the first part of the expression (!x || y) is false, but the second part !(!x || y) is true, therefore the whole expression is true.
  2. If both x and y have the initial value false, the expression will also evaluate to true. In this case, the first part of the expression (!x || y) will be true, and the second part !(!x || y) will also be true, resulting in the whole expression being true.
  3. If x has the value true and y has the value false, the expression will evaluate to true. Again, the first part of the expression (!x || y) will be true, and the second part !(!x || y) will also be true.
  4. In all other cases, the expression would not evaluate to true.

User Alen Paul Varghese
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