Question

Assume that x and y are boolean variables and have been properly initialized.(x && y) || !(x && y) The result of evaluating the expression above is best described as (A) always true (B) always false (C) true only when x is true and y is true (D) true only when x and y have the same value (E) true only when x and y have different values

Answer 1

The boolean expression (x && y) || !(x && y) is always true, due to the Law of Excluded Middle in boolean logic, making the correct answer (A) always true.

Step-by-step explanation:

The expression (x && y) || !(x && y) evaluates to a certain boolean value when x and y are boolean variables that have been properly initialized. In boolean logic, an AND operation combined with an OR operation on the same two variables, like in this expression, will always result in a true outcome, due to the Law of Excluded Middle which asserts that either a statement is true, or its negation is. This means that the statement inside the parentheses is either true (both x and y are true), or its negation outside the parentheses is true (not both x and y are true). Consequently, the correct answer is (A) always true.

Answer 2

(A) Always true

Step-by-step explanation:

Given

Boolean variables x and y

Required

What is (x && y) || !(x && y)

Irrespective of what x or y is, the x && y is always true

Take for instance:

x = true;

y = true;

x&&y will also be true because the values of x and y were not altered

And this is true for whatever boolean value x or y assume

Having said that:

x&&y = true

So, the expression is analysed as follows:

(x && y) || !(x && y)

Substitute true for x&&y

(true) || !(true)

!(true) = false

So, the expression becomes:

true || false

|| represents the OR operator; and it returns true if at least one of the conditions is true.

By this:

true || false = true

Option (A) Always true answers the question.