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.Q.Which of the following statements CANNOT be true ?a)P(x) = True for all x S such that x bb)P(x) = False for all x S such that x a and x cc)P(x) = False for all x S such that b x and x cd)P(x) = False for all x S such that a x and b xCorrect answer is option 'D'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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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.

Q. 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

Correct answer is option 'D'. Can you explain this answer?

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Let (5, ≤) be a partial order with two minimal elements a and b,...

'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.

Please comment below if you find anything wrong in the above post.

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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.Q.Which of the following statements CANNOT be true ?a)P(x) = True for all x ∈ S such that x ≠ bb)P(x) = False for all x ∈ S such that x ≠ a and x ≠ cc)P(x) = False for all x ∈ S such that b ≤ x and x ≠ cd)P(x) = False for all x ∈ S such that a ≤ x and b ≤ xCorrect answer is option 'D'. Can you explain this answer?

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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.Q.Which of the following statements CANNOT be true ?a)P(x) = True for all x ∈ S such that x ≠ bb)P(x) = False for all x ∈ S such that x ≠ a and x ≠ cc)P(x) = False for all x ∈ S such that b ≤ x and x ≠ cd)P(x) = False for all x ∈ S such that a ≤ x and b ≤ xCorrect answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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.Q.Which of the following statements CANNOT be true ?a)P(x) = True for all x ∈ S such that x ≠ bb)P(x) = False for all x ∈ S such that x ≠ a and x ≠ cc)P(x) = False for all x ∈ S such that b ≤ x and x ≠ cd)P(x) = False for all x ∈ S such that a ≤ x and b ≤ xCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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.Q.Which of the following statements CANNOT be true ?a)P(x) = True for all x ∈ S such that x ≠ bb)P(x) = False for all x ∈ S such that x ≠ a and x ≠ cc)P(x) = False for all x ∈ S such that b ≤ x and x ≠ cd)P(x) = False for all x ∈ S such that a ≤ x and b ≤ xCorrect answer is option 'D'. Can you explain this answer?.

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