By Gallier J.
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The problem with the above definition is that the universe of all possible expressions is not defined, and that the operations defined by (2) are not clearly defined either. This can be remedied as follows. Let Σ be the alphabet V ∪ {0, 1, (, ), +, ∗}, and A = Σ∗ be the set of all strings over Σ. The set A is the universe of all possible expressions. Note that A contains a lot of expressions that are not arithmetic expressions, and the purpose of the above inductive definition is to define the subset EXP R of Σ∗ describing exactly all arithmetic expressions.
We leave as an exercise the check that << is indeed a partial order on A × A. The following lemma will be useful. 3 If < A, ≤> is a well-founded partially ordered set, the lexicographic ordering << on A × A is also well founded. Proof : We proceed by contradiction. Assume that there is an infinite decreasing sequence (< xi , yi >)i∈N in A × A. Then, either, (1) There is an infinite number of distinct xi , or (2) There is only a finite number of distinct xi . In case (1), the subsequence consisting of these distinct elements forms a decreasing sequence in A, contradicting the fact that ≤ is well founded.
27 PROBLEMS (b) If N denotes the set of nonnegative integers and + is addition (of integers), show that there is some function h : X → N which does not have any homomorphic extension to A (a function g is a homomorphic extension of h if, g(b) = h(b) and g(x ∗ y) = g(x) + g(y), for all x, y ∈ A). Find an infinite set that is the inductive closure of a finite set, but is not freely generated. 2. Show that if X+ is freely generated by X and F , then Xin − Xi−1 = n (Xi − Xi−1 ) . Show that if X+ is not freely generated, then Xin − n Xi−1 = (Xi − Xi−1 )n is possible.