Question

Let P be the set of people we're interested in (not necessarily all people), Parent(p) mean p is a parent, and ParentOf(p1, p2) mean p1 is p2's parent. Consider the following statement: ∀x ∈ P, ∃y ∈ P, Parent(x) → ParentOf(x, y) If you were attempting to prove this statement via the challenge method, how would the "game" proceed? (An "adversary" is an opponent.) Group of answer choices If I can choose an x so that my adversary cannot choose a y that makes the statement false, then the statement is true. Otherwise, it's false. If I can choose an x and a y that make the statement true, then the statement is true. Otherwise, it's false. If my adversary cannot choose an x and a y that makes the statement false, then the statement is true. Otherwise, it's false. If no matter what choice of x my adversary makes, I can choose a y that makes the statement true, then the statement is true. Otherwise, it's false. None of these is correctly describes the process, but there is a correct process. None of these is correctly describes the process, and there is no correct process to prove this statement via the challenge method.

Answer 1

Answer:

If no matter what choice of x my adversary makes, I can choose a y that makes the statement true, then the statement is true. Otherwise, it's false.

Step-by-step explanation:

According to the given situation, as we can see that the we cant find the y for the x i.e given that shows Parent (x) represented as Parent Of (x,y)

In mathematically,

Parent (x) = Parent (x,y)

This does not seems true

Therefore this statement is false

Hence, the fourth option is true

All other statements that are mentioned is false