Logically Equivalent justify your answer

Question

1. (∀x (p(x) → q(x)) and (∀x p(x) → ∀x q(x))

2.∃x p(x)∧∃x q(x) and  ∃x (p(x)∧q(x))

3.(∀x (p(x) ↔ q(x)) and (∀x p(x) ↔ ∀x q(x))

are logically equivalent or not justify the answer

Comments

https://gateoverflow.in/306528/kenneth-rosen-edition-7th-exercise-1-4-question-43-page-no-56

Answer 1

1. (∀x (p(x) → q(x)) and (∀x p(x) → ∀x q(x))
     suppose domain consist of all positive integers
p(x) =x is divisible by 2 and q(x) =x is divisible by 4
we know that it is not always true that all positive integers are divisible by 2.
   consider ∀x p(x) → ∀x q(x),we know that ∀x p(x) is always false so this proposition must b true.
now consider ∀x (p(x) → q(x)) this proposition is false always bcoz it is saying that "if  a no x is divisible by 2 implies that it is divisible by 4".

2.∃x p(x)∧∃x q(x) and  ∃x (p(x)∧q(x))
     suppose domain consist of all positive integers
p(x)= x is odd   q(x)= x is even
    consider ∃x p(x)∧∃x q(x) ,it is saying that there exist odd numbers and there exist even numbers
now consider ∃x (p(x)∧q(x)) it is saying that there exist a no. which is is both odd and even it is trivially false.

3.(∀x (p(x) ↔ q(x)) and (∀x p(x) ↔ ∀x q(x))
     Let domain of x be 0 or 1, and
P(0) = false P(1) = true   Q(0) = true     Q(1) = false
Then we have: (P(0) = false) ≠ (Q(0) = true) which implies
∀x (p(x) ↔ q(x)) is FALSE.
On the other hand: P(0) = false, implies ∀xP(x) is false, and also Q(1) =
false, implies ∀xQ(x) is false, too. Consequently, (∀x p(x) ↔ ∀x q(x))​​​​​​​ is
TRUE.
conclusion that all of 3 are not logically equivalent.
 

Comments

P(x): x is prince.

Q(x):x has queen.

x: students in the class

(∀x (p(x) ↔ q(x)): If a boy is prince then he has queen.

(∀x p(x) ↔ ∀x q(x)) : If all students are prince then all of them has queen.

Are these two statement correct or I am making mistake?

Answer 2

Domain:a,b

1.(∀x (p(x) → q(x))  : (p(a)->q(a)) AND (p(b)->q(b))  

2. (∀x p(x) → ∀x q(x)) : p(a) AND p(b) -> q(a) AND q(b) since it is distributive over AND. Now using these simplified version, try to substitute truth values such that truth value of statement 1 IS NOT equal to statement 2 which is sufficient to prove that they r not equivalent.