self doubt

Question

Twin primes are pairs of numbers p and p+2 such that both are primes—for instance, 5 and 7, 11 and 13, 41 and 43. The Twin Prime Conjecture says that there are infinitely many twin primes. Let TwinPrime(n) be a predicate that is true if n and n+2 are twin primes. Which of the following formulas, interpreted over positive integers, expresses that there are only finitely many twin primes?

	∀m. ∃n. m ≤ n and not(TwinPrime(n))
	∃m. ∀n. n ≤ m implies TwinPrime(n)
	∃m. ∀n. TwinPrime(n) implies n ≤ m
	∀m. ∃n. n ≤ m and TwinPrime(n)

Comments

why it is not a

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Deepak%20Poonia sir can u expalin it little bit. https://math.stackexchange.com/questions/2920124/expressing-given-problem-in-first-order-logic

this is the actual question.

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See my answer here: https://gateoverflow.in/150941/self-doubt?show=425204#a425204 

Answer 1

Answer: $C$

Note that the domain is the set of positive integers. 

IDEA for CORRECT Translation:

We need finitely many TwinPrimes. So, there MUST be a number $m$, ahead of which there is No TwinPrime.

Also, If there is a number $m$, ahead of which there is No TwinPrime, then we don't have inifinitely many TwinPrimes, so, we are good. 

So, for finitely many TwinPrimes, we can say that there must be some $m$ such that for every $n \geq m$, $n$ is not TwinPrime. 

So, we get FOL expression: $\exists m \forall   n [ (n \geq m) \rightarrow Not(TwinPrime(n))  ] $

It can also be written as: $\exists m \forall   n [ (TwinPrime(n))  \rightarrow (n \lt m) ] $

Note that, the following are also correct, with same logic:

$\exists m \forall   n [ (n \gt m) \rightarrow Not(TwinPrime(n))  ] $

$\exists m \forall   n [ (TwinPrime(n))  \rightarrow (n \leq m) ] $

Option $A$ says that "for every positive integer $m$, there is some positive integer $n \geq m$ such that $n$ is not TwinPrime". It loosely says that "ahead of every number, there is some non-TwinPrime number".. But it doesn't prevent the case where we have infinite number of TwinPrimes. For example, IF (ASSUME) every odd number of TwinPrime then also option $A$ is true. 

Option $D$ says that "for every positive integer $m$, there is some positive integer $n \leq m$ such that $n$ is a TwinPrime". It loosely says that "before every number, there is some TwinPrime number".. But it doesn't prevent the case where we have infinite number of TwinPrimes. For example, IF (ASSUME) every odd number of TwinPrime then also option $D$ is true.

Option $B$ says that "there is some number $m$, such that every number $ \leq m$ is TwinPrime".. But it doesn't prevent the case where we have infinite number of TwinPrimes. For example, IF (ASSUME) all numbers are TwinPrime then also option $B$ is true.

Comments

yes sir now got it with clear understanding thank you.

Answer 2

https://math.stackexchange.com/questions/2920124/expressing-given-problem-in-first-order-logic Best answered with full explanation only !!!