ion is said to bind the free variable x in M . E.g. we say that x:yx has x as bound and y as free variable. Substitution [x := N ] is only performed in the free occurrences of x: yx(x:x)[x := N ] yN(x:x): In calculus there is a similar variable binding. In R b a f(x; y)dx the variable x is bound and y is free. It does not make sense to substitute 7 for x: R b a f(7; y)d7; but substitution for y makes sense: R b a f(x; 7)dx. For reasons of hygiene it will always be assumed that the bound variables that occur in a certain expression are dierent from the free ones. This can be fullled by renaming bound variables. E.g. x:x becomes y:y. Indeed, these expressions act the same way: (x:x)a = a = (y:y)a and in fact they denote the same intended algorithm. Therefore expressions that dier only in the names of bound variables are identied. 8 Introduction to Lambda Calculus Functions of more arguments Functions of several arguments can be obtained by iteration of applica...

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We use the Recursive Path Ordering (RPO) technique of semantic labelling to show the Preservation of Strong Normalization (PSN) property for several calculi of explicit substitution. Preservation of Strong Normalization states that if a term M is strongly normalizing under ordinary fi-reduction (using `global' substitutions), then it is strongly normalizing if the substitution is made explicit (`local'). There are different ways of making global substitution explicit and PSN is a quite natural and desirable property for the explicit substitution calculus. Our method for proving PSN is very general and applies to several known systems of explicit substitutions, both with named variables and with De Bruijn indices: AE of Lescanne et al., s of Kamareddine and R'ios and x of Rose and Bloo. We also look at two small extensions of the explicit substitution calculus that allow to permute substitutions. For one of these extensions PSN fails (using the counterexample in [Melli`es 95]). For the...

this paper is different and has been invented independently of the proofs in [Kamareddine & Rios 95] and [BBLR 95]. We show by means of a counterexample that an extension of exp with certain interaction between substitutions does not preserve strong normalisation. In appendix A we use a more common notation trying to determine the borderline between preservation of strong normalisation and interaction of substitutions. 2 The calculus

For a notion of reduction in a #-calculus one can ask whether a term satises conservation and uniform normalization. Conservation means that single-step reductions of the term preserve innite reduction paths from the term. Uniform normalization means that either the term will have no reduction paths leading to a normal form, or all reduction paths will lead to a normal form.