We can map t to this graph by applying the following linear tree homomorphism h into H 2 : A derivation of the form * 0 (t 1 , t 2 ) evaluates to the same graph as t 1 ; the graph value of t 2 is ignored.
then by lemma 11 again, n x , hence n t 0 x = n x and e t 0 x ≤ e t 0 x + n which yields Case 2 In this case Lemma 12 applies and at least one of C 2 -C 4 generate some token z ∈ {a, b, c, d}.