The Normalization Theorem
... (3) If t neutral and every time we convert a redex we get t0 ∈ R then t ∈ R. For atomic t, (1) is trivial. (2) holds because t is SN so all t0 are SN. For (3) we must show t is SN, so take any reduction step. You end up at an SN t0 so set ν(t) = ν(t0 ) + 1 and you’ve got t SN by the lemma. For t : U ...
... (3) If t neutral and every time we convert a redex we get t0 ∈ R then t ∈ R. For atomic t, (1) is trivial. (2) holds because t is SN so all t0 are SN. For (3) we must show t is SN, so take any reduction step. You end up at an SN t0 so set ν(t) = ν(t0 ) + 1 and you’ve got t SN by the lemma. For t : U ...
Motzkin paths and powers of continued fractions Alain Lascoux
... If we enumerate all Motzkin paths instead of the preceding ones, we have to correct by a factor `!/m1 ! m2 ! · · · which tells how many paths can be obtained from a given one by permutation of its irreducible components. Therefore one has : Theorem. For any n ∈ N, any k ∈ R , one has n ...
... If we enumerate all Motzkin paths instead of the preceding ones, we have to correct by a factor `!/m1 ! m2 ! · · · which tells how many paths can be obtained from a given one by permutation of its irreducible components. Therefore one has : Theorem. For any n ∈ N, any k ∈ R , one has n ...