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.

Question

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?

(A) P(x) = True for all x ∈ S such that x ≠ b

(B) P(x) = False for all x ∈ S such that x ≠ a and x ≠ c

(C) P(x) = False for all x ∈ S such that b ≤ x and x ≠ c

(D) P(x) = False for all x ∈ S such that a ≤ x and b ≤ x

In this question I am unable to get why we have a terminology of implication here as well as if some element is related to both a and b what will be its truth value, since P(a)=true and P(b)=False ,so If I have a maximal element c ,and it is related to both a and b then what will be its value ?

Answer 1

In this question I am unable to get why we have a terminology of implication here.

Because that is the problem you've been given. The idea is clearly to test your understanding of implications and of partial orders. Having separate concepts mixed together in problems is quite common. In fact, in application, it is difficult to find problems for which this is not the case.

if some element is related to both a and b what will be its truth value, since P(a)=true and P(b)=False

Well, if both $x > a$ and $x > b$, then $x > a$. So $P(a) \implies P(x)$. That is, $$\operatorname{True} \implies P(x).$$ That being the case, there is only one value $P(x)$ can be. This holds not just for $c$ but for every element related to $a$ (which are necessarily $\ge a$, since $a$ is minimal), whether it is related to $b$ or not.

If you understand implications, then you should recognize that elements related to $b$ are not restricted.

Answer 2

Firstly, as a point of correct use of MSE, your question should be self-contained. In particular we shouldn't have to read the title to find hypotheses for the question. Anyway the mathematics

Note that a lattice can't have two minimal elements, your tagging is incorrect.

So we have the following facts:

1. $a$ and $b$ are $\leq$-minimal.
2. $c$ is $\leq$-maximum.
3. $P(a)$ is true and $P(b)$ is false.
4. If $x \leq y$, then $P(x) \Rightarrow P(y)$.

From 2, we can conclude that $a \leq c$ and $b\leq c$. Also, as $a\neq b$ (because $P(a)$ is true and $P(b)$ is false, so, we know they are different!), and from 1, we know that $a < c$ and $b < c$.

Therefore, $c$ is an element satisfying $a \leq c$ and $b\leq c$.

By 4 and 3, we know that $P(c)$ is true. Therefore (D) cannot be true, as $c$ is something greater than both $a$ and $b$, and $P(c)$ is true.

If you take the partial order on $\{a,b,c\}$ given by $a \leq c$ and $b \leq c$, $P(a),P(c)$ are true and $P(b)$ is false, we have something satisfying all the hypotheses, and, also (A), (B) and (C), showing that these can be true.