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Let x = set of natural numbers.

EQUAL(X,Y) : 'X' is equal to 'Y'

EVEN(X) : 'X' is even.

PRIME(X) : 'X' is prime.

"The only even prime is "2"  can be expressed as :

∀x [ (EVEN(X) ∧ PRIME(X)) ↔ EQUAL(X,2)]

Here bi-direction is necessary because if only uni-direction implication were there from Left to Right and consider if number is even but prime ,so LHS becomes false so we can't say anything about RHS i.e whether number is two or not.This is so because in implication,if LHS is false we can't say anything about RHS and whole implication becomes true i.e F -> T or F is always true.

Hence we need both side implication to show that if given no. is both prime and even then it is only '2'  ,and if given no. is '2' then its even & prime both.

I hope it clear many doubts related to implication.

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I am not sure but you can try my option and let me know if i m wrong....

See Question is saying No. Should be even as well as prime(even and prime) and if it is true then no.has to be 2 otherwise formula sholud not exist....

If x={set of all integers} and

P(x):No.is even

Q(x):No.is Prime

R(x):No. is 2

then

for all x (P(x)^ Q(x)<--->R(x))

Double Implication bcz if no.is even and prime then it has to be 2 otherwise both has to be false

edited by
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∃x=2 A(x)

where A(x): x is even and prime

domain of x: set of all natural numbers

we can express this  statement with existential  quantifier  .
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