8 MIRNA DZAMONJAˇ no such choice is possible, stop the induction. Note that m ≤ k =⇒ γm ≤ γk and that the induction must be stopped at some stage k∗ as all ordinals that appear as unused at any stage come from the finite set γ∈dom(p)\{0} dom(fp(γ)) (and once an ordinal is used it remains so). Let γ∗ = γk∗. We shall now proceed to define a condition q in Nγ∗ ∩ γ∗. We shall need the following claim. Claim 3.4. For all k ≤ k∗, m≤k Nm ∩ γk ⊆ Nk. Proof of the Claim. The proof is by induction on k. If k = 0, the situation is clear. For the case of k + 1, we have by the induction hypothesis that for m ≤ k, Nm ∩ γk ⊆ Nk. Since γk+1 ≤ γk, certainly Nm ∩ γk+1 ⊆ Nk and hence it suﬃces to show that Nk ∩ γk+1 ⊆ Nk+1. This conclusion follows by the choice of Nk+1. 3.4 Suppose that k ≤ k∗ and m ≤ k. Then rm ∈ Nm so all the finitely many ordinals involved in the construction of rm γk are in Nm ∩ γk, and hence in Nk by Claim 3.4. Therefore rm γk ∈ Nk. At this point of the proof we shall get to use the Amalgamation Condition (∗), which we define in terms of a partially ordered class (K, ≤) of structures: (K, ≤) satisfies a.c. (∗) if for any elements A, B, {Ci : i ∈ I} of K satisfying • any two of these structures agree on their intersection • whenever i = j then either Ci ∩ Cj ⊆ A or Ci ∩ Cj ⊆ B, there is a structure in K which extends all A, B, {Ci : i ∈ I}. Obviously, the class of graphs with the induced subgraph relation satisfies the a.c. (∗) since we can simply take a union of all the graphs involved.4 Suppose now that β = β ∈ ak∗. By the induction hypothesis either Aβ ∩ Aβ δk∗, in which case sk∗ β ∩ sk∗ β ⊆ p(0) or Aβ ∩ Aβ ⊆ δk∗ and so sk∗ β ∩ sk∗ β ⊆ rk∗(0). This says that, according to the a.c. (∗) property, we can define a graph that extends p(0)∩δk∗ ∪rk∗(0)∪ β∈a k ∗ sk∗ β ∩δk∗. Let s be such a graph. We define q , a candidate for a condition in Pα. Let q (0) = s and for β ∈ k≤k∗ dom(rk)∪ak∗ let q (β) = p(β) ∩ δk∗ ∪ k≤k∗ rk(β). Note that q as a function is an element of Nk∗. We now claim that q is a condition in Pα. By induction on 1 ≤ β ≤ α we prove that q β ∈ Pβ. For β = 1 this follows from the choice of s and for β limit this is clear. Suppose that β = γ + 1 and γ is in dom(q ). If γ / dom(p) then γ ∈ k≤k∗ dom(rk) and q (γ) = k≤k∗ rk(γ). This is well defined by the choice of rk’s. If γ ∈ dom(p) \ k≤k∗ dom(rk), then q (γ) = p(γ) ∩ δk∗, so q β is a condition because p satisfies (i) in the definition of completeness. Finally it may happen that γ ∈ dom(p) ∩ k≤k∗ dom(rk). In particular we have β ∈ ak∗. Let k be maximal such that γ ≤ γk. We claim that dom(fp(γ)) ∩ δk∗ = dom(fp(γ)) ∩ δm. If this were not to be the case then at the stage k of the induction there would be an unused ordinal in dom(fp(γ)), so γ = γk+1, a contradiction. Having established the existence of q we now need to keep going to extend to a condition which extends each rk. This is done in stages using the assumptions of completeness, for which we refer the reader to the original article. 3.2 We now indicate how Lemma 3.2 implies that the forcing is ccc. We assume that we have proved that the set of complete conditions is dense, so we only work 4 In [9], Lemma 1, it is shown that a better known P(3−) amalgamation property implies a.c. (∗).

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