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generate candidate hypotheses hi that satisfy ( B A hi A x i ) I- f (xi),where B is the background knowledge plus any clauses learned on previous iterations Note this is an example-driven search, because each candidate hypothesis is constructed to cover a particular example Of course if multiple candidate hypotheses exist, then one strategy for selecting among them is to choose the one with highest accuracy over the other examples as well The CIGOL program uses inverse resolution with this kind of sequential covering algorithm, interacting with the user along the way to obtain training examples and to obtain guidance in its search through the vast space of possible inductive inference steps However, CIGOL uses first-order rather than propositional representations Below we describe the extension of the resolution rule required to accommodate first-order representations

At this stage I could say some things about the data model: A Client can appear in many journal entries, but an entry is for only one Client (Clients and journal entries therefore have a one-to-many relationship). A taskType can feature in many journal entries, but a journal entry has just one taskType (taskTypes and journal entries therefore are also one-to-many). Actually, a single journal entry could contain more than one taskType, but that would complicate things. If I design my model and implement it correctly, this is something that I could change later. I decided to keep it simple. A journal entry is worked on by one Consultant, while a Consultant will have many entries, so Consultants and journal entries have a one-tomany relationship. The Consultant can work for any Client. However, the relationship is not significant in the data model until I introduce multi-Consultant features, and those features aren t going in yet, so I don t need to create a relationship. A Consultant needs to record the charge for a journal entry. Here is my refined list of features: I can create, edit, and delete journal entries. I can create, edit, and delete Clients. I can create, edit, and delete task types. At this point, I could sketch out a data model (see Figure 9 2):

The resolution rule extends easily to first-order expressions As in the propositional case, it takes two clauses as input and produces a third clause as output The key difference from the propositional case is that the process is now based on the notion of unifying substitutions We define a substitution to be any mapping of variables to terms For example, the substitution 6 = {x/Bob, y / z } indicates that the variable x is to be replaced by the term Bob, and that the variable y is to be replaced by the term z We use the notation WO to denote the result of applying the substitution 6 to some expression W For example, if L is the literal Father(x, Bill) and 6 is the substitution defined above, then LO = Father(Bob, Bill) We say that 6 is a unifying substitution for two literals L1 and L2, provided LlO = L2O For example, if L1 = Father(x, y ) , L2 = Father(Bil1, z ) , and O = ( x / B i l l ,z / y } , then 6 is a unifying substitution for L1 and L2 because LlO = L2O = Father(Bil1, y) The significance of a unifying substitution is this: In the propositional form of resolution, the resolvent of two clauses C1 and C2 is found by identifying a literal L that appears in C 1 such that -L appears in C2 In firstorder resolution, this generalizes to finding one literal L1 from clause C1 and one literal L2 from C2, such that some unifying substitution 6 can be found for L1 and -L2 (ie, such that LIO = -L20) The resolution rule then constructs the resolvent C according to the equation

The general statement of the resolution rule is shown in Table 107 To illustrate, suppose C 1 = White(x) t Swan(x) and suppose C2 = Swan(Fred) To apply the resolution rule, we first re-express C1 in clause form as the equivalent expression C 1 = White(x)v -Swan(x) The resolution rule can now be applied In the first step, it finds the literal L1 = -Swan(x) from C1 and the literal L2 = Swan(Fred) from C2 If we choose the unifying substitution O = { x / F r e d } then these two literals satisfy LIB = -L20 = -Swan(Fred) Therefore, the conclusion C is the union of (C1 - {L1})O= White(Fred) and (C2 - {L2})0= 0, or C = White(Fred)

That model also suggests something about the user interface, doesn t it Good application design ensures that you handle your data in the appropriate place. This data model has the journal entry as the centerpiece, and it makes sense to have this data available in the main window. Likewise it seems like a good idea to manage Clients, Consultants, and tasks as secondary and separate windows. Figure 9 3 is an updated sketch for the user interfaces you are going to build. It s still not very detailed it doesn t have to be.