By Andrzej Grzegorczyk
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Extra resources for Fonctions Recursives
This only depends on γi for pi ≤P pj . If such an extension exists, put Ak (x) = A 1 − Φe j (x). If there is no such extension, do nothing. Then, go to stage s + 1. To verify that the construction satisﬁes all the requirements, for Rk,j,e consider the stage s = k, j, e . Either we extended some γ or we did not. A If we extended some γ, then there is x such that Φe j (x) ↓= Ak (x). If we A did not, then no such extension exists and since Aj extends γj,s , Φe j (x) ↑. 10: Every countable partial order can be embedded in D.
9 The direct extension ordering ≤∗ on Qn is deﬁned to be ≤ 0 ∪ ≤1 . 10 Let p, q ∈ Qn . Then p ≤ q iﬀ either (1) p ≤∗ q or (2) p = a, A, f ∈ Qn0 , q ∈ Qn1 and the following holds: (a) (b) (c) (d) q⊇f dom(q) ⊇ a q(max(a)) ∈ A for every β ∈ a q(β) = πmax(a),β (q(max(a))). e. the Cohen forcing. However, the following basic facts relate it to the Prikry type forcing notion. 22 M. 1. 11 Qn , ≤∗ is κn -closed. e. for every p ∈ Qn and every statement σ of the forcing language there is q ≥∗ p deciding σ.
8 Qn = Qn0 ∪ Qn1 . 9 The direct extension ordering ≤∗ on Qn is deﬁned to be ≤ 0 ∪ ≤1 . 10 Let p, q ∈ Qn . Then p ≤ q iﬀ either (1) p ≤∗ q or (2) p = a, A, f ∈ Qn0 , q ∈ Qn1 and the following holds: (a) (b) (c) (d) q⊇f dom(q) ⊇ a q(max(a)) ∈ A for every β ∈ a q(β) = πmax(a),β (q(max(a))). e. the Cohen forcing. However, the following basic facts relate it to the Prikry type forcing notion. 22 M. 1. 11 Qn , ≤∗ is κn -closed. e. for every p ∈ Qn and every statement σ of the forcing language there is q ≥∗ p deciding σ.