Answer : Only S2 satisfies given formula $\psi.$ Hence, None of the options in correct.
Wait, I know you might not agree with me, but read the whole answer first.
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.
The notation a | b means that a divides b.
Hence, $5$ is divisible by $1,5,-1,-5.$ $1$ is divisible by $1,-1.$ 1 divides everything. So does −1.
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."
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, $-w$. 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..)
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.
Clearly, S2 satisfies $\psi.$ Neither $S1,S3$ satisfies $\psi.$
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