To resolve this question, we will consider all the options one by one and try to make the LHS (Left Hand Side) TRUE and RHS (Right Hand Side) FALSE;
We’ll consider a domain x with two values: x1 and x2
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X
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P(x)
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Q(x)
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X1
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T
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F
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X2
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F
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T
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- ((∀x(P(x) v Q(x)))) → ((∀(x)P(x) v (∀(x)Q(x)))
(P(x1) v Q(x1)) ^ (P(x2) v Q(x2)) → (P(x1) ^ P(x2) v (Q(x1) ^ Q(x2))
(T v F) ^ (F v T) → (T ^ F) v (F ^ T)
(T) ^ (T) → (F) v (F)
(T) → (F)
(F)
In this case, in the LHS, at least one of the predicates of the domain x1 and of the domain x2 must be TRUE for the LHS be true. In the RHS, for all x, either all the values in P(x) are TRUE (P(x1)) and (P(x2)), or all the values in Q(x) are TRUE (Q(x1)) and (Q(x2)). It’s impossible because (P(x1) ^ P(2)) is FALSE and (Q(x1) ^ Q(2)) is also FALSE.
We were able to make LHS TRUE and the RHS FALSE. So, as what will remain is an implication between TRUE → FALSE, the result will be FALSE.
- (∀x(P(x) → Q(x))) → ((∀x(Px) → (∀xQ(x)))
(Px1) → Q(x1)) ^ (P(x2) → Q(x2)) → (P(x1) ^ P(x2)) → (Q(x1) ^ Q(2))
(T) → (F) ^ (F) → (T) → (T) ^ (F) → (F) ^ (T)
(F) ^ (T) → (F) → (F)
(F) → (T)
(T)
In this case, in the LHS, for the implication (P(x)) → (Q(x)) be TRUE, either both P(x) and Q(x) are TRUE or at least Q(x) must be TRUE. However, this turns out to be impossible because of the domains x1 and x2. (P(x1) → Q(x1)) and (P(x2) → Q(x2)) will result in the conjunction “FALSE and TRUE”, which in turn will result in FALSE.
In the RHS, we have an implication of all the values of P(x) in Q(x). Therefore, as the outcome of (P(x1) ^ P(x2)) and (Q(x1) ^ Q(x2)) will be FALSE, we are able to make the RHS FALSE.
Thus, we end up having an implication of FALSE to FALSE, resulting in TRUE.
- (∀x(P(x)) → ∀x(Q(x))) → (∀x(P(x)) → Q(x)))
(P(x1) ^ P(x2)) → (Q(x1) ^ Q(x2)) → (P(x1) → Q(x1)) ^ (P(x2) → Q(x2))
(T) ^ (F) → (F) ^ (T) → (T) → (F) ^ (F) → (T)
(F) → (F) → (F) ^ (F)
(T) → (F)
(F)
In this case, we have the exact opposite present in option B. This will culminate in a implication of TRUE to FALSE, which will result FALSE.
- (∀x(P(x)) ↔ (∀x(Q(x)))) → (∀x(P(x) ↔ Q(x)))
(P(x1)) ^ P(x2)) ↔ (Q(x1)) ^ Q(x2)) → (P(x1) ↔ Q(x1)) ^ (P(x2)) ↔ (Q(x2))
(T) ^ (F) ↔ (F) ^ (T) → (T) ↔ (F) ^ (F) ↔ (T)
(F) ↔ (F) → (F) ^ (F)
(T) → (F)
(F)
In this case we have a double implication in the LHS and in the RHS. In the LHS, for every value of P(x), we have a implication of these values in Q(X) (for both domains x1 and x2). We will have as a result a double implication between FALSE and FALSE, which will result in TRUE.
In the RHS, we have a double implication of the predicates of P and Q for both domains x1 and x2, which will result in: (P(x1) ↔ Q(x1)) ^ (P(x2) ↔ Q(x2)). The result will be a conjunction of FALSE and FALSE, which in turn will result in FALSE.
Finally, we will have a implication of TRUE to FALSE, which will result in FALSE.
Therefore, option B is the correct.
