Question

Given that A and B are true and X and Y are false, determine the truth value of the following proposition: ~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ ~X) v (B ⊃ X)]

Answer 1

Answer:

The value of the proposition is FALSE

Step-by-step explanation:

~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ ~X) v (B ⊃ X)]

Let's start with the smallest part: ~X. The symbol ~ is negation when X is true with the negation is false and vice-versa. In this case, ~X is true (T)

~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ T) v (B ⊃ X)]

Now the parts inside parenthesis:  (A ⊃ Y),(X ⊃ B),(A ≡ T) and (B ⊃ X). The symbol ⊃ is the conditional and A ⊃ Y is false when Y is false and A is true, in any other case is true. The symbol ≡ is the biconditional and A ≡ Y is true when both A and Y are true or when both are false.

(A ⊃ Y) is False (F)

(X ⊃ B) is True (T)

(A ≡ T) is True (T)

(B ⊃ X) is False (F)

~[(F) v ~(T)] ⋅ [~(T) v (F)]

The two negations inside the brackets must be taken into account:

~[(F) v F] ⋅ [F v (F)]

The symbol left inside the brackets v is the disjunction, and A v Y is false only  with both are false. F v (F) is False.

~[F] ⋅ [F]

Again considerating the negation:

T⋅ [F]

Finally, the symbol ⋅ is the conjunction, and A v Y is true only with both are true.

T⋅ [F] is False.