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Which of the following are true :

1. $\forall x (P(x) \Leftrightarrow Q(x) ) \rightarrow (\forall x P(x) \Leftrightarrow \forall x Q(x))$

2. $\forall x (P(x) \Leftrightarrow Q(x) ) \leftarrow (\forall x P(x) \Leftrightarrow \forall x Q(x))$

3. $\forall x (P(x) \Leftrightarrow Q(x) ) \Leftrightarrow (\forall x P(x) \Leftrightarrow \forall x Q(x))$

4. $\exists x (P(x) \Leftrightarrow Q(x) ) \rightarrow (\exists x P(x) \Leftrightarrow \exists x Q(x))$

5. $\exists x (P(x) \Leftrightarrow Q(x) ) \leftarrow (\exists x P(x) \Leftrightarrow \exists x Q(x))$

6. $\exists x (P(x) \Leftrightarrow Q(x) ) \Leftrightarrow (\exists x P(x) \Leftrightarrow \exists x Q(x))$

1 Answer

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Only $1$ is True... Rest of all are False.

Let me show any Two of the above with Formal proof..And rest can be solved similarly.

1.  ∀x(P(x)⇔Q(x))  →   (∀xP(x)⇔∀xQ(x)) : ($A \rightarrow B$)

It is an Implication statement of the form $A \rightarrow B$...Hence It is Always True except when $A$ is True and $B$ is False.

So, Instead of proving any implication statement True, We should usually  try to prove it False as Any Implication statement will be False in only one case.

Similarly, here also, We would make $A$ True and then we will Try to make $B$ false. If we can somehow do it...then we can say that the implication $A \rightarrow B$ is False. And If we can never make $A$ True and $B$ false Simultaneously then  we can say that the implication $A \rightarrow B$ is True.

So, Let's make $A$ true i.e. $\forall x(P(x) \Leftrightarrow Q(x))$ True.

$x$ belongs to some Non-empty domain. So, Let me take Three element ($x1,x2,x3$) in the domain..(Note that Domain need not have only or exactly three elements..there could be infinite elements as well...But with experience I found that for such equivalences(distributive properties) in logic, every type of equivalence can be shown True/False with accuracy by just having $3$ elements in the domain.. )

So, Now, In order to make $A$ True i.e. $\forall x(P(x) \Leftrightarrow Q(x))$ True for $x1,x2,x3$..We need to make $(P(x) \Leftrightarrow Q(x)$ True for all the Three elements. Hence, There could be several different possibilities to do that... We need to find at least one such possibility in which $A$ becomes true and $B$ becomes False.

For $x1$ : $P1 \Leftrightarrow Q1$... Either both $P1,Q1$ can be True or both $P1,Q1$ can be False....(To make $P1 \Leftrightarrow Q1$ True for $x1$ )

For $x2$ :$P2 \Leftrightarrow Q2$... Either both $P2,Q2$ can be True or both $P2,Q2$ can be False..

For $x3$ :$P3 \Leftrightarrow Q3$... Either both $P3,Q3$ can be True or both $P3,Q3$ can be False..

So, there could be Two cases :

  • Either All $Pi, Qi$ can be True ..But in this case, $B$ can not become False.
  • Or At least one of $Pi,Qi$ is False...Let it be $P2$..i.e. $P2 = F$..But as we know $P2$ and $Q2$ has to be same in order to make $A$ True..Hence, $Q2 = F$ as well. And In this case also, You can check $B$ will be True..

So, We can never make $A$ true and $B$ false simultaneously. Hence, The Implication  ∀x(P(x)⇔Q(x))  →   (∀xP(x)⇔∀xQ(x)) Is Valid and True.


Let me show one more with same idea :

5 :  ∃x(P(x)⇔Q(x))   ←   (∃xP(x)⇔∃xQ(x)) ..

Rewriting it : $(\exists xP(x) \Leftrightarrow \exists x Q(x)) \rightarrow \exists x(P(x) \Leftrightarrow Q(x))$ .....--- ($A \rightarrow B$)

Applying Same idea here as well :

$x$ belongs to some Non-empty domain. So, Let me take Three element ($x1,x2,x3$) in the domain..

So, Now, In order to make $A$ True i.e. $(\exists xP(x) \Leftrightarrow \exists x Q(x)) $ True..Either Both $\exists xP(x)$ and $\exists x Q(x)$ will be True or Both will be false.

So, Let's make them both True by having $P1 = T$ and $Q2 = T$..Hence, now $A$ is True.

So, Now we will try to make $B$ false...For that, We can have $Q1 = F$, $P2 = F$, $P3 = T$, $Q3 = F$...Hence, After making $A$ true, Now we have simultaneously made $B$ false. Hence, the implication $(\exists xP(x) \Leftrightarrow \exists x Q(x)) \rightarrow \exists x(P(x) \Leftrightarrow Q(x))$ is Not Valid or True. 


NOTE 1 : This is a Formal approach to prove these distributive equivalences True/False. Hence, In first glance it might look hard or confusing But if done a lots of practice of this approach then It becomes really easy to answer any such equivalence given to you in any exam.

NOTE 2 : Having $3$ elements in the domain is a conclusive result that I have found with practice and experience..,, to accurately say True/False for such Implications.

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