Construction of H. The chain of inclusions Y ⊆ L(H ) ⊆ L(G) is sufficient to prove Lemma 6: Because Y and L(G) are Parikh-equivalent (which we observed at the beginning of Section 3.4.2), so are L(H ) and L(G), which means that L(H ) satisfies all of the properties claimed in Lemma 6, even though this does not suffice to prove our current lemma.
once H is given, it is not hard to also obtain a grammar H that generates exactly Y.