@Digvijay Pandey
That means for every x there exist a w which is greater than x and it is prime.
No, this is not the correct interpretation for the given predicate formula.
This interpretation will become false for the following set : $Set \,\,of\,\, all\,\, even\,\, \,\, numbers \,\, greater \,\, than \,\, 2 $ i.e. $ S = \{ 4,6,8,...... \}$
For set $S$ above, the interpretation "for every $x$ there exist a $w$ which is greater than $x$ and it($w$) is prime" will return false. But for this set $S$, given predicate formula $\psi$ will be True.
Correct interpretation :
Given predicate formula says "For every number $x$ in the domain, if $x$ is such that $x$ is divisible by only $1,x$ among all the elements in the whole domain , then there exists some number $w > x$ in the domain such that $w$ is divisible by only $1,w$ among all the elements in the whole domain.
In all the other answers to this question, the interpretation that is given is "for every prime number(x) there exist another prime number(y) where x<y."
This interpretation is WRONG.
Why is this interpretation wrong ?
Divisibility Definition : If a and b are integers, a divides b if there is an integer c such that ac = b.
Hence, $5$ is divisible by $1,5,-1,-5.$ $1$ is divisible by $1,-1.$ 1 divides everything. So does −1.
Standard(General) definition of Prime : "An integer n > 1 is prime if its only positive divisors are 1 and itself.
The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, . . . .
An integer n > 1 is composite if it isn’t prime."
http://sites.millersville.edu/bikenaga/number-theory/divisibility/divisibility.pdf
https://primes.utm.edu/glossary/page.php?sort=Divides
Now, Take the following set as domain for the given formula $\psi$ : Set of All integers.
For Set of ALL integers, the given formula $\psi$ DOES NOT satisfy. Because for element $-1$ in the domain, $\psi$ will become false. See, for $x=-1$, first part will become true i.e. for $x =-1,$ (∀z z∣x ⇒ ((z=x)∨(z=1))) will become true because only $1,-1$ divides $-1$. But for $x=-1,$ 2nd part will become false. i.e. For $x=-1,$ There does not exist any number $w > -1$ in the domain such that that $w$ is divisible by only $w,1$ because every number is also divisible by -1. So, for $x = -1,$ $\psi$ becomes false. Hence, $\psi$ is NOT satisfied by the set "Set of All integers."
Hence, Answer should be Only S2. But none of the options match (Doesn't mean that correct answer will change because options do not match..)
Moreover, in the answer(above), you have divided the given formula in three parts. But it is not so. 2nd and 3rd part are not individual but it makes one part as a whole. $\exists w$ in the 2nd part is applied on whole remaining formula i.e. the scope of $\exists w$ is the whole remaining formula (to the right).
So, the given formula is basically $\forall x [1 \rightarrow 2]$
where $2$ is : $ \exists w ( w > x) \wedge (\forall z \: z \mid w \Rightarrow ((w=z) \vee (z=1))) $
Here $1$ says "$x$ is a divisible by only 1,$x$ in the whole doamin"
$2$ says "There exists a $w > x$ which is also divisible by only 1,$w$ in the whole domain "
So, whole formula interprets as "For every number $x$ in the domain, if $x$ is divisible by only $1,x$ among all the elements in the whole domain , then there exists some number $w > x$ in the domain such that $w$ is divisible by only $1,w$ among all the elements in the whole domain.
@Akanksha Agrawal
Whatever you have read, is not wrong. But whenever an author writes a book, he describes some conventions that he'll be using throughout the book and so all the examples and questions in that particular book should be interpreted as per those conventions. So, what you said is a convention followed by Keneth rosen, in order to use less number of parentheses or brackets to make the expression look nice and not too much crowded.
But different authors may opt for different conventions. Here in the question, I think they have followed the convention that the scope of a quantifier is as long as you can take, until a closing parentheses occurs for that quantifier. (this convention was used in one of the courses I took in IISc.. Professor in the beginning described that this convention he'll be following).
Like in the part $1,$ $"( \forall "$ means that now until you find a closing parentheses to this opening parentheses, it will be scope of $\forall z$ quantifier. Similarly, For $\exists w$ , there is no opening parentheses for $\exists w$ because they intend to make its scope the whole remaining expression, so they did not need to put opening parentheses for $\exists w.$
Ideally, in GATE exam, they should give the expression in proper parentheses.
Most of the times, you will find GATE questions to be properly framed.
Now, How do we know that the quantifiers in this question has the scope as described above and not according to the convention followed by Kenneth rosen ?
That's because Assume that you follow Kenneth rosen's conventions then $\exists w$ will have scope only til $(w > x) $ i.e. $\exists w (w > x)$ is one whole part and $\exists w$ won't have any scope any further. But then what is $w$ in the remaining part i.e. $ (\forall z \: z \mid w \Rightarrow ((w=z) \vee (z=1))) $ ??
Here, in the remaining part $w$ will be a free variable and hence, we CAN NOT interpret the given formula $\psi$ until/unless we are provided with the interpretation of $w.$ Similarly for $\forall z$ in the part $1$ and $forall z $ in the part 2.
I'd again say that in GATE most questions are framed unambiguously and this question should have been framed in a better way by providing proper parentheses....But one can "understand the convention" followed by the question setter by looking at the formula. So, Technically there is no ambiguity But still it should have been framed in better way.
Refer my answer here for few more details :
https://gateoverflow.in/302813/gate2019-35?show=328427#a328427
