By Gallier J.

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**Example text**

Later on when we define satisfaction in first-order logic, we will need to define the concept of a structure, and we will have to reformulate slightly the notion of an inductive closure. This can be done conveniently by introducing the concept of an algebra. Since this material is only used in Chapter 5 and Chapter 10, it has been included in an appendix. 1. Show the following properties: (a) If there exists a function g : B → A such that f ◦ g = IA , then f : A → B is injective. If f : A → B is injective and A = ∅, then there exists a function g : B → A such that f ◦ g = IA .

The bottom-up method is to define a sequence EXP Ri of subsets of A by induction as follows: EXP R0 = V ∪ {0, 1}; EXP Ri+1 = EXP Ri ∪ {H+ (u, v), H∗ (u, v)|u, v ∈ EXP Ri }, for i ≥ 0. We let X+ = EXP Ri . We shall show below that X+ = X + and therefore, EXP R is equal to X+ . 1, we give the following general definition. Let A be a set, X ⊂ A a subset of A, and F a set of functions f : An → A, each having some arity n > 0. , yn ) is also in Y . Clearly, A itself is inductive on X. The intersection of all inductive sets on X is also closed under F and it is called the inductive closure of X under F .

For each y in Y , choose some x in X such that y ≺ x, and add a successor node labeled y to the node corresponding to that x. For every remaining x in X (element that is dropped and replaced by no elements at all), add a successor labeled with the special symbol ⊥. This last step guarantees that at least one new node is added to the tree for every multiset Mn in the sequence. This is necessary in case Y is empty. Repeat the process for M1 >> M2 , M2 >> M3 , and so on. Let T be the resulting tree.