GATE | GATE-CS-2003 | Question 31

Let (5, ≤) 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 ?
(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

Explanation:
‘a’ and ‘b’ are given as minimal elements. No other element in S is of lower order than either a or b.
‘c’ is given as maximum element. So, c is of higher order than any other element in S.

P(a) = True means all elements ‘x’ which have an edge from element ‘a’ have to be true.
Since there is an edge from ‘a’, we have to satisfy formula P(a) => P(x), which can only be done by setting
P(x) = True.

Elements which have an edge from b can be anything because formula P(b) => P(x) is satisfied as P(b) = False.

(A) This statement is true because making all elements true trivially satisfy formula P(x) => P(y).

(B) This statement is true if all elements are connected from b then all elements can be false.

(C) This statement is true because b<=x ensures x!=a and for all other elements P(x) can be false without violating the given implication.

(D) This statement is false. Since, P(a) = true , for all ‘x’ such that a<=x, P(x) must be true. We do have at least one such 'x', which is 'c' as it is the maximum element.

Thus, option (D) is the answer.

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