This is the easiest question in the world if you understand what it says.
two minimal elements a and b, and a maximum element c.
This means that no other element relates to a and b, while every element relates to c. (Research on maximal, maximum, minimal and minimum yourself)
So, the partial order looks like this.
Where these crisscrossed lines represent any network that leads from a to c, and b to c.
We know that a and b are not related, because minimal. But c essentially relates to both.
Let P: S → {True, False} be a predicate defined on S.
Now we've attached a predicate to our POSET. Of course, it can have only two values — True or False.
Suppose that P(a) = True, P(b) = False and P(x) ⟹ P(y) for all x,y∈S satisfying x≤y
With that Predicate, we gave truth values to a and b. And for any element x related to y, P(x) —> P(y) (implication statement)
So, our Partial Order can look like this now.
By virtue of implications in Predicate Logic, True —> True, but False —> True or False. (And therefore c is definitely true!)
Now lets come to the options. Btw Option D is incorrect straightforward because a relates to x, and a is True so x can't be false. But let's see in details.
- P(x) = True for all x ∈ S such that x ≠ b
If x is not b, x can be a. And a is True. So this option can be true.
- P(x) = False for all x ∈ S such that x ≠ a and x ≠ c
If x is not a or c, we can say x is b. So x is False, and this option can also be true.
- P(x) = False for all x ∈ S such that b ≤ x and x ≠ c
b is related to x. And x isn't c (c is 100% true, see below) So, x can be False. Because False —> True or False both.
- P(x) = False for all x ∈ S such that a ≤ x and b ≤ x
a and b are related to x. Whatever a relates to has to be True. Whatever b relates to can be anything. So when both relate to some element, it has to be True. So this option is incorrect
PS: And hence c is also True.