Show that two quantified statements are logically equivalent

[Mr Davis 97]

Homework Statement

Show that $\forall x(P(x) \land Q(x)) \equiv \forall xP(x) \land \forall xQ(x)$

Homework Equations

The Attempt at a Solution

Based on my work from propositional logic, to show that two expressions are logically equivalent, then we have to show that $\forall x(P(x) \land Q(x)) \Longleftrightarrow \forall xP(x) \land \forall xQ(x)$ is a tautology; that is, it is always true. It is always true if they have the same truth values for all x in the domain. For propositional logic, it was a matter of listing out the finite combinations of truth values and showing that they are always the same. However, with predicate logic, we are dealing with an infinite domain of discourse, so we can't just list them off. How should I proceed then?

 

[Stephen Tashi]

Show that $\forall (P(x) \land Q(x)) \equiv \forall P(x) \land \forall Q(x)$

What variable does $\forall$ quantify? Did you mean $( \forall x, P(x) \land Q(x)\ ) \equiv (\forall x, P(x))\land(\forall x, Q(x))$.

What assumptions and theorems have you studied in proposition logic ? "Universal generalization", "Existential instantiation" etc. ?

 

[Mr Davis 97]

What variable does $\forall$ quantify? Did you mean $( \forall x, P(x) \land Q(x)\ ) \equiv (\forall x, P(x))\land(\forall x, Q(x))$.

What assumptions and theorems have you studied in proposition logic ? "Universal generalization", "Existential instantiation" etc. ?

I fixed the errors in the original post.

I was thinking that maybe I could use universal generalization to show that, if the LHS is true, then $P(a) \land Q(a)$ is true for all a in the domain. Then by universal generalization, we would have the RHS. However, I don't see how this establishes that the biconditional is a tautology, which is necessary to show that they are logically equivalent.

 

[Stephen Tashi]

However, I don't see how this establishes that the biconditional is a tautology, which is necessary to show that they are logically equivalent.

A biconditional is just two conditionals. Show each separately.

 

[Mr Davis 97]

A biconditional is just two conditionals. Show each separately.

If I show that LHS implies the RHS, and that the RHS implies the LHS, then I would show that one being true implies that the other is true. But in order for it to be a tautology, don't you have to also show that if one is false than the other must be false as well?

 

[Stephen Tashi]

If I show that LHS implies the RHS, and that the RHS implies the LHS, then I would show that one being true implies that the other is true. But in order for it to be a tautology, don't you have to also show that if one is false than the other must be false as well?

That will depend on how your text materials define the relation "$\equiv$".

Is there a theorem or definition in your materials that says: $( ( A \implies B) \land (B \implies A) ) \implies A \equiv B$ ?

 

[fresh_42]

But in order for it to be a tautology, don't you have to also show that if one is false than the other must be false as well?

It is always a good idea to start this way: What does it mean, if one side were wrong? Can the other still be true?
And then the other way around.