@debasree88

P(x)=¬(x=1)∧∀y(∃z(x=y∗z)⇒(y=x)∨(y=1))

Let focus on second part :- **B = ∀y(∃z(x=y∗z)⇒(y=x)∨(y=1))**

if x = prime (suppose 5) **B will be false if there exit atleast one y for which ∃z(x=y∗z)⇒(y=x)∨(y=1) is false**.

now check for y = 1,2,3,4,5,6 (or any Positive integer)

{ if y = 1 then Z will be 5 hence B become (T--->T) which results in True.}

{if y = 2 then Z does not exist hence B become (F--->F) which results in True.}

{if y = 3 then Z does not exist hence B become (F--->F) which results in True.}

{if y = 4 then Z does not exist hence B become (F--->F) which results in True.}

{if y = 5 then Z will be 1 hence B become (T-->T) which results in True.}

and so on

So by observing above scenario **“ if x is prime then B is True for all value of y”.**

if x = not Prime (suppose 4)** B will be false if there exit atleast one y for which ∃z(x=y∗z)⇒(y=x)∨(y=1) is false.**

now check for y = 1,2,3,4,5,6 (or any Positive integer)

{ if y = 1 then Z will be 4 hence B become (T--->T) which results in True.}

**{ if y = 2 then Z will be 2 hence B become (T--->F) which results in False.}**

now without checking further we can say that B is False Hence P(x) is false for x = composite number.

Please correct me if I am wrong!