CS 380/480 - Foundations of Artificial Intelligence - Winter 2007

Assignment 2
Sample Problems and Solutions


  1. Consider the sentence F: (A /\ B) => (B \/ C)

    (a)  Is this sentence satisfiable? If so, how many models does it have?

    ANSWER:

    A sentence is satisfiable if there is at least one interpretation that makes it true. For propositional sentences such as the one in this problem, we can determine this by constructing the truth table for the sentence and see if there is one assignment of truth values to the variables that makes this sentence true (in other words, if we find a row in the truth table which results in the value "true" for the sentence). The truth table below shows the intermediate columns for different sub-sentences in F with the final column corresponding to the full sentence. The first 3 columns represent the different combinations of truth value assignments to the individual variables A, B, and C in the sentence. [Note that there are 8 = 23 possible truth value assignments because we have 3 variables. In general, the truth table would have 2n rows, where n is the number of propositional variables in the sentence].
     
    A B C B A /\ B C B \/ C (A /\ B) => (B \/ C)
    false false false true false true true true
    false false true true false false false true
    false true false false false true true true
    false true true false false false true true
    true false false true true true true true
    true false true true true false false false
    true true false false false true true true
    true true true false false false true true

    Note that the two intermediate columns for A /\ B and B \/ C represent the two sides of the implication in the full sentence F. By definition of =>, F would be true if the left-hand-side is false or if the right-had-side is true. This gives the final column for F in the above truth table.

    An interpretation for the sentence F is any truth value assignment for the variables A, B, and C (a row in the truth table). Clearly the sentence F is satisfiable because it is true in at least one interpretation (i.e. it has at least one model). The models for F are all the interpretations that make the sentence true. In this case we see that F has 7 models.

    It is also worth noting that F is only false in one interpretation (namely, A=true, B=false, C=true). If F had been true in all of the 8 possible interpretations, then it would have been a valid sentence (a.k.a., a tautology). In this case, F is not valid, because it is not true in all interpretations.

    (b) Is the sentence B entailed by F (i.e., does it logically follow from F)?

    ANSWER:

    To answer this type of question, we must construct truth tables for F and for B. The sentence B is entailed by F (written as F |= B), if any interpretation that makes F true, also makes B true. In other words, any model of F should also be a model of B (note, however, that models of B need not be models of F).

    In this case we can easily observe (form the above truth table) that there are several interpretations that make F true, but for which B is false. For example, the interpretations A=false, B=true, C=false (3rd row in the truth table), is a model for F, but is not a model for B. Therefore, B is not entailed by F (does not logically follow from F).

    To see an example of a situation where one sentence entails another, consider the sentence B \/ C. We can observe in the above truth table that any interpretation that makes B \/ C true also makes F true. Therefore, B \/ C entails F (or, F logically follows from B \/ C). We can express this using the notation: (B \/ C) |= F.

  1. Problem 7.4 on Page 237 of Russell and Norvig.

    2. In this problem you are asked to "prove" various assertions that reflect different properties of the entailment relationship ( |= ). To prove these assertions, you need to give an argument based on the definitions of entailment and various logical operations. Here we consider parts (a) and (c) of this problem as examples of what you need to do.

    (a) a sentence a is valid, if and only if True |= a.

    ANSWER:

    Because this is an "if and only if" assertions, we need to argue in both direction. So, we must show that if  a is valid, then True |= a; and conversely, we must also show that if True |= a, then a is valid.

    First let us assume that a is valid. By definition of validity, a must be true in every interpretation. In particular, is true in every interpretation that makes any other sentence b true. In other words, if a is valid then it is entailed by any other sentence b (including the universal constant True which by itself is a sentence). Therefore, True |= a.

    This shows the assertion in the "if" direction. Now, we need to also show the "only if" direction. To do so, let us assume that True |= a. This would mean that a is true in every interpretation that makes the sentence True also true (i.e., every model of True is also a model of a). But, by definition, the constant True is always true in every interpretation. Therefore, every interpretation is also a model for a. This, by definition of validity means that a is a valid sentence.

    (c) a |= b if and only if the sentence (a => b) is valid.

    ANSWER:

    As in the previous case, we must show both direction.

    "If" Direction:  Assume that a |= b. By definition of entailment, this means any model of a is also a model of b. Now, suppose that I is any arbitrary interpretation. If I makes a true, then by the entailment assumption it would also make b true, in which case (a => b) is true by the definition of implication operator. On the other hand, if I makes a false, then regardless of b, by the definition of implication, (a => b) is true. So, we can conclude that (a => b) is true under any interpretation, and hence, by definition of validity, it is a valid sentence.

    "Only If" Direction:  Let us assume that the sentence a => b is valid, in other words it is true in every interpretation. We want to show that any interpretation that makes a true, will also make b true. Suppose that I is an arbitrary interpretation that makes a true (i.e., I is an arbitrary model of a). Since by our assumption a => b is valid, then I must also be a model of a => b (i.e., a => b is true in I). Since a is true in I, then b must also be true in I, or else a => b would be false, contradiction our validity assumption. Thus, if we assume that is a => b valid, then any model of a is also a model of b. In other words, a |= b.

    This completed the proof of the assertion.

     

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