Resolution for Predicate Logic

A substitution is a function θ from the set of σ-terms back to itself such that (writing function application on the right) cθ = c for each constant symbol c and f(t1, . . . , tk)θ = f(t1θ, . . . , tkθ) for each k-ary function symbol f . It is clear that the composition of two such substitutions (as functions) is also a substitution. We have previously considered substitutions of the form [t/x] for a σ-term t and variable x. We write composition of substitutions diagrammatically, that is, θ · θ′ denotes the substitution obtained by applying θ first and then θ′. (This convention matches the fact that for substitutions we write function application on the right.) In particular [t1/x1] · · · [tk/xx] denotes the substitution obtained by sequentially applying the substitutions [t1/x1], . . . , [tk/xk] left-to-right. Given a set of literals D = {L1, . . . , Lk} and a substitution θ, define Dθ := {L1θ, . . . , Lkθ}. We say that θ unifies D if Dθ = {L} for some literal L. For example, the substitution θ = [f(a)/x][a/y] unifies {P (x), P (f(y))}, as does the substitution θ′ = [f(y)/x]. In this example we regard θ′ as a more general unifier because θ = θ′ · [a/y], that is, θ factors through θ′. We say that θ is a most general unifier of a set of literals D if θ is a unifier of D and any other unifier θ′ factors through θ, i.e., we have θ′ = θ · θ′′ for some substitution θ′′. Note that both the substitutions [x/y] and [y/x] are both most general unifiers of {P (x), P (y)} (in fact most-general unifiers are only unique up to renaming variables). We will show that a set of literals either has no unifier or it has a most general unifier. Examples of sets of literals that cannot be unified are {P (f(x)), P (g(x))} and {P (f(x)), P (x))}. The problem in the second case is that we cannot unify a variable x and term t if x occurs in t.