Sets and quantifiers question

Question

Am I doing this correctly?

Let S be a non-empty set, and let P(x) and Q(x) be open sentences that can be applied to any x∈S. For each of the following implications, determine whether or not it is true for every possible S, P(x), Q(x). If so, give a brief explanation. If not, give a counterexample of S, P(x), Q(x) for which it is false.

a) If (∀x ∈ S, P(x)) and (∀x ∈ S, Q(x)), then ∀x ∈ S, (P(x) and Q(x)).

My answer: False. Disprove by counterexample.

S:{1,2,3}

P(x): x >/= 3

Q(x): x < 0

x ∈ S does not satisfy P(x) because {1,2}

x ∈ S does not satisfy Q(x) because {1,2,3}

since neither P(x) or Q(x) is satisfied ∀x ∈ S, statement is false.

b) If (∃x ∈ S, P(x)) and (∃x ∈S, Q(x)), then ∃x ∈ S, (P(x) and Q(x)).

False. Disprove by counterexample.

S: {1,2,3}

P(x): x >/= 3

Q(x): x < /= 1

There exists an element in set S that satisfies P(x), {3}, but it does not satisfy Q(x). Likewise {1} satisfies Q(x) but does not satisfy P(x). Therefore since no element in S satisfies P(x) and Q(x) statement is false.

That's what I have. I'm not sure if it's correct since I'm new to this. Can you guys correct me if I made a mistake somewhere?

Answer 1

For part (a) the statement is true in your example, because it has the form $F\Rightarrow F$, which is a true statement - revise the truth table for implication if you are not clear on this.

IMHO, "non-mathematical" examples are often best for thinking about logic. So how about this: Let $S$ be the set of all students in your logic class, let $P(x)$ mean "$x$ is under $20$ years old" and let $Q(x)$ mean "$x$ has blue eyes". Supposing (I'm not saying it's really true, but just supposing) you know that everyone in the class is under $20$ years old, and you know that everyone in the class has blue eyes. Can you say for certain that everyone in the class is under $20$ and has blue eyes?

If you can say this for certain, and if you can briefly explain why, then you have an answer to your question.

In part (b) I think you have a typo because your specification of $Q(x)$ does not make any sense, however if you fix this it looks to me as if you are heading in the right direction.

Answer 2

Edit @OP:

Question A is of the form:

$$\forall xP(x)\wedge\forall xQ(x) \rightarrow \forall x(P(x)\wedge Q(x))$$

You can determine that there is no counterexample by showing that the above statement is a tautology. By your answer:

$$P(x) \equiv F, Q(x) \equiv F$$ $$(\forall x P(x) \wedge \forall x Q(x)) \rightarrow \forall x (F \wedge F)$$ $$(F \wedge F) \rightarrow F$$ $$F \rightarrow F$$ $$T$$

It might be helpful to add that question a is an exercise to show the idea that

$$\forall xP(x)\wedge\forall xQ(x) \equiv \forall x(P(x)\wedge Q(x))$$

as David mentioned the statement is of the form $$p\rightarrow q$$ where $$p \equiv q$$ so the conditional statement is $$p \rightarrow p$$ which is a tautology $$p \rightarrow p \equiv T$$ so there is no counterexample to show that the conditional statement is false.