Question 9611 – Propositional-Logic

α: (∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒ (∀x)[¬Px]
I1: Domain: the set of natural numbers
    Px ≡ 'x is a prime number'
    Qxy ≡ 'y divides x'
I2: same as I1 except that Px = 'x is a composite number'.

Given that:
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Q_yy]] ⇒(∀x)[¬Px]
Q_yy is always true, because y divide y, then ¬Q_yy is false.
∀x[(P(x) ⇔ ∀y [Qxy ⇔ False]]
∀y [Qxy ⇔ False] can be written as ∀y[¬axy]

⇒(∀x)[P(x) ⇔ ∀y[¬Qxy]]
Here, ¬Qxy says that y doesnot divides x, which is not always be true.
For example, if x=y then it is false then ∀y[¬Qxy] is not true for all values of y.
⇒(∀x)[P(x) ⇔ False]
⇒(∀x)[¬P(x) = RHS]
LHS = RHS
⇒ Irrespective of x, whether x is prime of composite number I₁ and I₂ satisfies α.