**Ans Option D**

This problem seems confusing to many Lets make it more simple::

First what is "**satisfiability**"::::Meaning can you find **a case where it is true ;**

So our goal should to find A condition where above expression is **TRUE**;

So now come to format of question.

IT is having** A**$\Rightarrow$**B** which means we have three way to make it true(using TT of implication)

Come To *I2
As Px is prime no so For composite no Px will be false That means Compliment Of Px will be TRUE for every case(*

*As here universal quantifier)*

**So right site of implication is true so it will be True for all case.SO****VALID**

*come to I1:*

**(**∀x)[Px⇔(∀y)[Qxy⇔¬Qyy]]⇒**(∀x)[¬Px] **

As **Px is prime no so right side .. compliment of Px will be False now In implication If right side is false then for To be True Left also should be false. One thing we have to focus now that how can we make left For atleast one Case **

So when we come inside the Bi-implication ([**Px⇔(∀y)[Qxy⇔¬Qyy]]) **we have Px at one side which will be always true As Px is prime no.So to make biconditional false right side we have to make false . Again right side is having biconditional **[Qxy⇔¬Qyy]]**

Since we know ¬Qyy will always be fase because Qyy will always True (as every no will divide with itself).

So

**Qxy::meaning that y divide x .**

** as we know when Y= 1and Y=x(meaning no itself) it will divide so Qxy True .**

So for **[Qxy⇔¬Qyy]] we have two case where it will evalate False.(using TT)**

** for this Px⇔(∀y)[Qxy⇔¬Qyy]] right side is false in two cases and left will always be true.overall it will be false(universal quantifier).**

** (∀x)[Px⇔(∀y)[Qxy⇔¬Qyy]]**⇒(∀x)[¬Px]

So for above expression we shown that Left is false for two cases and Right will always be False So It give us True for

some cases so It is also **Valid**>

**ANS OPTION D**