One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included in its positive closure (that is, L² ⊆ L⁺). The intuitive idea is that if we had a counterexample for closure under concatenation that uses just a single language L, then if there was some pair of strings in L² that invalidates suffx substitution closure-that yields a string not in L² when the suffx of one is substituted into the other-then that pair would invalidate suffx substitution closure for L* as well. But this argument doesn't work. The fact that the pair yields a string that is not in L₂ does not rule out the possibility of string being in Li for some i = 2.

If one thinks in terms of strictly local generation, it should be clear that a language L is strictly 2-local language i? it includes all and only the strings that start with a symbol from some particular subset of Σ and end with a symbol from another such subset, with only  particular pairs of adjacent symbols occurring in between-equivalently, some particular set of forbidden pairs not occurring (see Section 3 of Part 1).

Consider, then L⁺. Strings in L⁺ will also start and end with symbols from those subsets of Σ and the adjacent pairs of symbols occurring strictly within the string from a given iteration of L will be only those that are permitted. The only di?erence is that there may be additional adjacent pairs where the strings from successive iterations meet. These we can admit by simply permitting them as well. The question is whether they will allow pairs in the middle of a string from L which should be forbidden. But, since we are only adding pairs in which the left symbol is a permissible ending symbol for a string from L and the right symbol is a permissible starting symbol, everywhere such a pair occurs is a permissible boundary between strings of L. Finally, to extend the construction to get L* all we need to do is add the pair ?? as well.