Sets And Relation
Question 1 
Consider the binary relation R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE?
R is symmetric but NOT antisymmetric
 
R is NOT symmetric but antisymmetric
 
R is both symmetric and antisymmetric
 
R is neither symmetric nor antisymmetric

Question 1 Explanation:
Symmetric Relation: A relation R on a set A is called symmetric if (b,a) € R holds when (a,b) € R.
Antisymmetric Relation: A relation R on a set A is called antisymmetric if (a,b)€ R and (b,a) € R then a = b is called antisymmetric.
In the given relation R, for (x,y) there is no (y,x). So, this is not Symmetric. (x,z) is in R also (z,x) is in R, but as per antisymmetric relation, x should be equal to z, where this fails.
So, R is neither Symmetric nor Antisymmetric.
Antisymmetric Relation: A relation R on a set A is called antisymmetric if (a,b)€ R and (b,a) € R then a = b is called antisymmetric.
In the given relation R, for (x,y) there is no (y,x). So, this is not Symmetric. (x,z) is in R also (z,x) is in R, but as per antisymmetric relation, x should be equal to z, where this fails.
So, R is neither Symmetric nor Antisymmetric.
Question 2 
Question 2 Explanation:
π_{4} = refines every partition. So it has to be bottom of poset diagram.
And, neither π_{2} refines π_{3}, nor π_{3} refines π_{2}.
Here, only π_{1} refined by every set, so it has to be at the top.
Finally, option C satisfies all the property.
And, neither π_{2} refines π_{3}, nor π_{3} refines π_{2}.
Here, only π_{1} refined by every set, so it has to be at the top.
Finally, option C satisfies all the property.
Question 3 
Let (S, ≤) be a partial order with two minimal elements a and b, and a maximum element c. Let P: S → {True, False} be a predicate defined on S. Suppose that P(a) = True, P(b) = False and P(x) ⇒ P(y) for all x, y ∈ S satisfying x≤y, where ⇒ stands for logical implication. Which of the following statements CANNOT be true?
P(x) = True for all x ∈ S such that x ≠ b
 
P(x) = False for all x ∈ S such that x ≠ a and x ≠ c  
P(x) = False for all x ∈ S such that b ≤ x and x ≠ c  
P(x) = False for all x ∈ S such that a ≤ x and b ≤ x

Question 3 Explanation:
c is the maximum element.
a or b the minimal element in set.
P(a) = True for all x ∈ S such that a ≤ x and b ≤ x.
Option D is False.
a or b the minimal element in set.
P(a) = True for all x ∈ S such that a ≤ x and b ≤ x.
Option D is False.
There are 3 questions to complete.