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Which one of the following options is CORRECT given three positive integers $x, y$ and $z$, and a predicate
$$P\left(x\right) = \neg \left(x=1\right)\wedge \forall y \left(\exists z\left(x=y*z\right) \Rightarrow \left(y=x\right) \vee \left(y=1\right) \right)$$

  1. $P(x)$ being true means that $x$ is a prime number
  2. $P(x)$ being true means that $x$ is a number other than $1$
  3. $P(x)$ is always true irrespective of the value of $x$
  4. $P(x)$ being true means that $x$ has exactly two factors other than $1$ and $x$
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P(x)=¬(x=1)∧∀y(∃z(x=y∗z)⇒(y=x)∨(y=1))   here a simple way to solve it is  according to options  A. P(x) being true means that x is a prime number.  put here x=1 in LHS which is TRUE so negation of this will be FALSE and false logical and with anything will be false so LHS of implication() is false and we know that if LHS of implication is false then we don't care about RHS because F⇒  ANYTHING WILL give TRUE.  so 

A is right option.

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lets break this in simple terms

first x cant be 1

now for all y given (

if there is some z such that x=y*z then y=1 or y= x)

now take x=4, then for all y(

take y=1, then some z is 4 i.e 4=1*4 so T->T T

take y=4, then some z is 1 i.e. 4=4*1 so T->T T

take y=2,then some z is 2 i.e.4=2*2 so T->F F as y != 2 or x(4) so it is not true for all y hence x= 4 is eliminated in fact all composite will be false and only prime no will be true for all y because for e.g 7=1*7 or 7=7*1 both cases satisfy this for all y and other than 1 and 7 you cant find any other factors .

so x represents all prime no and 1 is not prime as it is already removed at start.
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