$\forall x(P(x) \rightarrow Q(x))$ and $(\exists xP(x) \rightarrow \forall yQ(y))$ having different truth values

Question

I've just been introduced to predicate logic and have been stuck on this question:

Define predicates $P(x)$ and $Q(x)$ with the same domain such that the formulas $$\forall x(P(x) \rightarrow Q(x))$$ and $$(\exists x\,P(x)) \rightarrow (\forall y\,Q(y))$$ have different truth values. Justify your answers briefly.

Does the first statement say that for all $x$, $P(x)$ implies $Q(x),$ and does the second statement say that there exists an $x$ such that if $P(x)$ is true then $Q(x)$ is true?

What is the intuition for coming up with the required answer? I've tried predicates such as "$x$ is even" or "$x$ is prime", and haven't come up with a satisfactory answer. Maybe my understanding is flawed.

Answer 1

$$\forall \color\red x\,\Big(P(\color\red x) \rightarrow Q(\color\red x)\Big)\tag1$$ and $$\Big(\exists \color\red x\,P(\color\red x)\Big) \rightarrow \Big(\forall \color{blue}y\,Q(\color{blue}y)\Big)\tag2$$

Does the second statement say that there exists an $x$ such that if $P(x)$ is true then $Q(x)$ is true?

In statement $(2),$ notice that the scope of $x$ is the antecedent (the 'if' part) while the scope of $y$ is the consequent (the 'then' part). Thus, statement $(2)$ can actually be rewritten as $$\Big(\exists \color\red x\,P(\color\red x)\Big) \rightarrow \Big(\forall \color{blue}x\,Q(\color{blue}x)\Big),\tag2$$ and says that

• if $P$ is true of some object, then $Q$ is true of every object. $\qquad(2)$

This is a stronger assertion than statement $(1),$ which merely says that

• if $P$ is true of some object, then $Q$ is true of that same object. $\qquad(1)$

I've tried predicates such as "$x$ is even" or "$x$ is prime"

Letting the domain of discourse be $\mathbb Z$ and defining $P(x)$ and $Q(x)$ both as "$x$ is even", what are the truth values of statements $(1)$ and $(2)$ ?

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Side note: statement $(2)$ is in fact equivalent to $$\forall \color\red x\,\Big(P(\color\red x) \rightarrow \forall \color{blue}y\,Q(\color{blue}y)\Big).\tag2$$ This shows a starker constrast with statement $(1):$ $$\forall \color\red x\,\Big(P(\color\red x) \rightarrow Q(\color\red x)\Big).\tag1$$

Answer 2

Thinking gives me a headache, and I only do it as a last resort. What I do with a problem like this is just try easy examples at random and if I'm lucky one of them will be a counterexample. If it was a numerical identity to disprove, I'd try plugging in $x=0,1,2$. With your logic problem, we could start by assuming that $P$ and $Q$ are the same, $Q=P$. In that case the two formulas are $$\forall x(P(x)\to P(x))\text{ and }(\exists xP(x)\to\forall yP(y)).$$ Now what? The formula on the left is going to be true. The one on the right could be true, but there are lots of ways it could go wrong. Can you take it from here?

Answer 3

The first statement says that $P$ implies $Q$ for every $x$. The second statement says that as long as one $x$ satisfies $P$ then all $y$ satisfy $Q$.

For example, let $P(x)$ be the statement "x is from Dallas" and $Q(x)$ be the statement "x is from Texas". Then $$\forall x(P(x)\to Q(x))$$ says "for all persons, if they are from Dallas then they are from Texas". $$(\exists x P(x))\to \forall y Q(y)$$ says "If there is a person from Dallas then everyone is from Texas".