Please help me in regard with this question.I didn't have a clue how to solve this.
The way I thought about this question is assuming the truth values of predicates P(x) and Q(x) and then trying different combinations.
Determine whether following propositions are equivalent. Justify your answers.
- ∀x (P(x) → Q(x)) and ∀x (P(x)) → ∀x (Q(x)).
- ∀x (P(x) ⇔ Q(x)) and ∀x (P(x)) ⇔ ∀x (Q(x)).
Please provide a logical approach, not a mathematical approach. Thank You.
The first proposition in the first question says:
The second says:
To see whether two propositions are equivalent, it’s often helpful to ask under what circumstances each of them would be false. The first one would be false if there were some $x$ that had property $P$ but not property $Q$. The second would be false only if every $x$ had property $P$, but not every $x$ had property $Q$. It’s pretty clear that these are different requirements. Imagine that the universe of discourse contains just two objects, $a$ and $b$. Object $a$ has property $P$, object $b$ does not have property $P$, and neither $a$ nor $b$ has property $Q$. In this tiny system the first proposition is false, because $P(a)$ is true and $Q(a)$ isn’t. The second proposition, however, is true, because it’s an implication whose antecedent, $\forall xP(x)$, is false. (Remember that $p\to q$ is always true when $p$ is false.)
If you want a more ‘natural’ example, let the domain of discourse be the positive integers, let $P(x)$ be ‘$x$ is odd’, and let $Q(x)$ be ‘$x$ is prime’. In this model the first statement is false, since $15$ is odd but not prime. The second, however, is vacuously true: it says that if every positive integer is odd, then every positive integer is prime, and since it’s certainly not the case that every positive integer is odd, the proposition is automatically true. (If you’re at all bothered by this, ask yourself what evidence would be required to demonstrate conclusively that the implication was false: you’d need to show that even though every positive integer was odd, there was some composite positive integer. Since you can’t even show that every positive integer is odd (since it’s not true), you certainly can’t show that every positive integer is odd and some positive integer is composite.)
Now let’s look at the second question. The first proposition says:
The second says:
Suppose that the first proposition is true. There are two cases to consider: either every $x$ in the universe of discourse has property $P$, or there is at least one that does not. In the first case the truth of the proposition means that every $x$ in the universe of discourse also has property $Q$. Thus, every object in the universe of discourse has both properties, as asserted by the second proposition. In the second case there is at at least one object that does not have $P$, and the truth of the first proposition ensures that it also fails to have $Q$. Thus, $\forall x(P(x))$ and $\forall x(Q(x))$ are both false, a situation that is also consistent with the truth of the second proposition. In short, if the first proposition is true, so is the second:
$$\forall x(P(x)\leftrightarrow Q(x))\;\to\;\big(\forall x(P(x)\leftrightarrow Q(x))\big)\;.$$
But the second proposition can be true even if the first is false. Suppose that our universe of discourse contains just two people: Tripta, who is female and single, and Bhushan, who is male and married. Let $P(x)$ be ‘$x$ is female’, and let $Q(x)$ be ‘$x$ is married’. In this tiny model the first proposition is false: it says that each of these two people is female if and only if that person is married, and Bhushan is a counterexample: he is married and not female.
The second proposition, however, is true. It’s not the case that everyone in the universe of discourse is female, so $\forall xP(x)$ is false, and the second proposition then says that $\forall xQ(x)$ must be false as well. And indeed it is: Tripta is not married, i.e., $Q(\text{Tripta})$ is false.
Thus, the two propositions are not equivalent.