Chapter11 Flashcards
What is Clarks Idea?
The idea of program completion is to turn such implications into a definition by adding the
corresponding necessary counterpart
Every stable model of P is a model of CF(P)?
Yes But. not vise versa, a stable model for the completion of P is not a always a stable model of P.
Models of the completion of P are supported models.
What causes the mismatch between supported
and stable models?
Answer The mismatch between supported and stable models
is caused by cyclic derivations
Fages’ Theorem
Let P be a tight normal logic program and X ⊆ A(P)
Then, X is a stable model of P iff X |= CF(P)
For L ⊆ A(P), the external supports of L for P are?
ESP(L) = {r ∈ P | h(r) ∈ L, B(r)+ ∩ L = ∅} External supports for L are defined as a rule that exists in P or the head of a rule that exists in the positive rule body wholse intersection with the loop is the empty set.
what is B(r)+?
The positive rule body.
What is |=?
The right hand side is the logical consequence of the left hand side.
How do you read x = 3?
x is assigned the value 3 and not as x is 3
If a set of atoms X is a model of a normal program P, then X is a model of the completion of P.
False. X would need to be a stable model to be a model of the completion. models of the completion are supported models, one of them is the stable model.
If a set of atoms X is a stable model of a normal Program P, then X is a model of the completion of P.
True.
If a set of atoms X is a model of the completion of P, then it is a model of the normal program P.
True
If a set of atoms X is a model of the completion of P, then it is a stable model of the program P.
FALSE. It is supported.
Given a normal program P, if the completion of P has at least one model, then P has at least one stable model.
False.
Let P be a normal program and X a set of atoms such that X |= CF(P).
If X is not a stable model of P, then there is a loop L in P such that L is contained in X and X satisfies the loop formula P(L)
False
Let P be a normal program and L1. L2 are elements of loop(P) if L1 intersect L2 in not the empty set then L1 union L2 are both elements of loop(P)
False.