Question

boolean x = true;
boolean y = false;
boolean z = true;

Given the following declarations, evaluate each boolean expression labelled a-f below as either true or false: (Section 3.7)

a. y && z

b. y || z

c. !y

d. (x && y) || (x && z)

e. !(x && y)

f. (!y) && (x && z)

Answer 1

boolean x = true;

boolean y = false;

boolean z = true;

a. y && z

Answer: y && z -> False && true ->false

The above statement is resolved as false && true, the logical and (&&) returns true if both the operands are true, here in our case both the operands are not true hence the expression returns false.

b. y || z

Answer: y || z -> false || true -> true

The above statement is resolved as false || true, the logical or (||) returns true if any one of the operands is true, here in our case one of the operands is true hence the expression returns true.

c. !y

Answer: true

In the above statement logical not (!) is used which gives the reverse of the result after evaluating the expression. Now, we have y=false, hence !y return true.

d. (x && y) || (x && z)

Answer: (true && false) || (true && true) -> (false) || (true)-> true

For the above expression the answer is true, here we have applied OR between two expressions.

The first expression (x && y) returns false since y is false

The second expression (x && z) returns true since both x and z are true

Finally applying Or between the above two we have false|| true which returns true

e. !(x && y)

Answer: !(x && y) -> !(true && false) -> !(false) -> true

In the above expression (x && y) is false as y is false, now we have !(false) which is true

f. (!y) && (x && z)

Answer: (!y) && (x && z) -> !(false) && (true && true) -> !(false) && (true) -> true && true

-> true

In the above statement we have && applied between two expression

The first expression !y returns true as y is false

The second expression x && z returns true as both x and z are true.

Now (!y) && (x && z) is resolved as true && true which is true