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Question

Consider the following formula and its two interpretations $I1$ and $I2$.

$α:(∀x)[Px⇔(∀y)[Qxy⇔¬Qyy]]⇒(∀x)[¬Px]$

$I1 :$ Domain: the set of natural numbers

$Px =$ $\text{'x is a prime number'}$

$Qxy =\text{ 'y divides x'}$

$I2 :$ same as $I1$ except that $Px = \text{'x is a composite number'}.$

Which of the following statements is true?

$\text{I1 satisfies α, I2 does not}$
$\text{I2 satisfies α, I1 does not}$
$\text{Neither I1 nor I2 satisfies α}$
$\text{Both I1 and I2 satisfies}$

just one doubt
when it says $p(x): x \text{ is a prime number}$
$α:(∀x)[Px⇔(∀y)[Qxy⇔¬Qyy]]⇒(∀x)[¬Px]$

then, in that $(∀x)$ does that means $x$ itself is a set of prime numbers and for all $x$ quantifier is going to pick every prime number value from the set

when it says $p(x): x \text{ is a composite number}$
$α:(∀x)[Px⇔(∀y)[Qxy⇔¬Qyy]]⇒(∀x)[¬Px]$

then, in that $(∀x)$ does that means $x$ itself is a set of composite numbers and for all $x$ quantifier is going to pick every composite number value from the set

and the domain of $Y:\text{{set of all natural numbers}}$

Comments

No.
P(x) is just a preposition, which says x is a prime number.
The domain is set of natural numbers. Both $x,y\  \epsilon \ \mathbb{N}.$
$\forall x $ means for all natural number x.

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which means ∀ (x),∀ (y) in that everytime x,y is going to pick random numbers from set of natural numbers

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Yes.

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thank you so much @Soumya29

Answer 1

https://gateoverflow.in/923/gate2003-33

this link will provide the answer for the question