@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!