By Irène Guessarian (auth.)

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**Example text**

Example. >.. >.. f (g(v,g(h(v),G(h2(v)))), G(v)) > ... it will g e n e r a t e the language f(L(S,G(~)),~) generate any tree t' such trees n e v e r t h e l e s s Hence, outside-in f(tl,t 2) with Q ~ t2~L(S,G(~)) ; belong to L(S,t). although any tree in L(S,t) computation OI c o m p u t a t i o n of the form but will fail to sequence, we may can be obtained by some as well, by applying some sequence,fail to obtain~/l the trees we wish. will be e x p r e s s e d by saying that the correct /DS, MNV,V/.

T"' lemmas To this end, w e a p p l y t' c M(F,V) To p r o v e Let then c l e a r l y t S* case, ~ " - - > as in the p r e v i o u s Exercise:Try for some t i in T i easily: whence Y(S) is the least fixpoint of S. the reverse inclusion, S* t in T i, t' in L i s u c h that t - - > t' and Q = ~, . Then: t' (ZI~) = t' ~ t(~* B u t S(~) c i t(L /G) c ÷ ~ T i (L/G) the theorem. implies for L c Y(S). is a f i x p o i n t To p r o v e ~ k (~) = ~k(~) (Z) i ~). c i and S(~) = S(~)i We h a v e ~*(~) c ~ .

Later - Proofs can be found in /BO, BL, DO, F, The shortest ones are in /N2,BL/ ; h o w e v e r both need some m i n o r a d j u s t m e n t s in order to be totally accurate. , n, Qi is a subset of M(F,{v I .... ,Vr(Gi)}). , n, Qi c Qi! where . e. any subset of C has a least u p p e r b o u n d in C). 35: Let S be the mapping (where T = (T 1 ..... 30 QG (continued): {a , h(v) couple: ({a,g(v,a) , g(v,a)} G = (GI, .... Gn)) Let ' actually Q e (QG,QH) with: QH = { a , g(h(v),u)} ; then , g(v,h2(v)) , g(v,g(h(v),a))} The reader is strongly advised others) from C to C defined by: by writing to check this the t e r m s i n a tree-like the substitutions.