SUBSECTOR TELECOMUNICACIONES

             W

a∈{uj| uj∈ui}ψ(a) if b ≡ ui,

W suppL X(a)< fin Xsψ(a) if b ≡ L u s, W suppL X(a)< fin Xsθ(a, ~c) ∧ ψ(a) if b ≡ {y ∈ Lu_s| θ(y, ~c)}, ∃_xψ(x) 'W a∈Lu Xψ(a).

Observe that these disjunctive clauses cover precisely one of the formulas ϕ and ¬ϕ, for each Lu

X-formula ϕ. In view of ϕ ≡ ¬¬ϕ we may assume that ϕ itself is the

disjunctive formula. Then the formula ¬ϕ is covered by the stipulation ¬ϕ 'V

a∈ιX(ϕ)¬ ϕa if ϕ '

W

a∈ιX(ϕ)ϕa.

In other words, if the formula ϕ is disjunctive then the formula ¬ϕ is conjunctive and we have ιX(¬ϕ) = ιX(ϕ) as well as (¬ϕ)a≡ ¬(ϕa) for a ∈ ιX(ϕ).

Let us show that the verification calculus is natural:

Lemma 3.1.14. Given any order embedding f : X → Y , the LuX-formula ϕ and

the Lu

Y-formula ϕ[f ] have the same type (disjunctive or conjunctive), we have

a ∈ ιX(ϕ) ⇔ Luf(a) ∈ ιY(ϕ[f ])

for all a ∈ Lu_X, and

ϕ[f ]_Lu

f(a)≡ ϕa[f ]

holds for all a ∈ ιX(ϕ).

Proof. First assume that ϕ is disjunctive. We look at some representative cases: Consider a formula of the form

ϕ ≡ b ∈ ui,

where b is not of the form uj, and observe that we have

ϕ[f ] ≡ Lu_f(b) ∈ ui.

The function Lu_f is injective, as it is an order embedding by Lemma 3.1.10. Also note that Lu

f is the identity on u. This implies that Luf(b) is still not of the form uj,

so that both ϕ and ϕ[f ] are disjunctive. We also get

a ∈ ιX(ϕ) ⇔ a ∈ {uj| uj ∈ ui} ⇔ Luf(a) ∈ {uj| uj ∈ ui} ⇔ Luf(a) ∈ ιY(ϕ[f ]), as well as ϕ[f ]Lu f(a)≡ L u f(a) = Luf(b) ≡ (a = b)[f ] ≡ ϕa[f ].

Next, assume that we have

ϕ ≡ b ∈ {x ∈ Lu_s| θ(x, c1, . . . , ck)}

and thus

ϕ[f ] ≡ Lu_f(b) ∈ {x ∈ Lu_{f (s)}| θ(x, L_fu(c1), . . . , Luf(ck))}.

Both formulas are disjunctive. As f is an order embedding and suppL_· is a natural transformation we can deduce

a ∈ ιX(ϕ) ⇔ suppLX(a) <finX s ⇔ [f ]<ω(suppLX(a)) <finY f (s) ⇔

⇔ suppL_Y(Lu_f(a)) <fin_Y f (s) ⇔ Lu_f(a) ∈ ιY(ϕ[f ]).

Since θ(a, c1, . . . , ck) does only contain the displayed parameters we also see

ϕa[f ] ≡ Luf(a) = Luf(b) ∧ θ(Luf(a), Luf(c1), . . . , Luf(ck)) ≡ ϕ[f ]Lu f(a).

Concerning the cases ϕ ≡ b0 6= b1 and ϕ ≡ ϕ0 ∨ ϕ1 we remark that Lu_f is the

identity on {0, 1} ⊆ u. As a last example, assume that we have ϕ ≡ ∃xψ(x)

and thus

ϕ[f ] ≡ ∃xψ[f ](x).

Both formulas are disjunctive. In view of ιX(ϕ) = LuX and ιY(ϕ[f ]) = LuY we have

a ∈ ιX(ϕ) and Lu_f(a) ∈ ιY(ϕ[f ]) for all terms a ∈ Lu_X. Above we have observed

that substitution is natural. Using this fact we get

ϕa[f ] ≡ ψ(a)[f ] ≡ ψ[f ](Luf(a)) ≡ ϕ[f ]Lu f(a),

as required. Having checked all disjunctive cases, assume now that ϕ is a con- junctive formula. Then ϕ is of the form ϕ ≡ ¬ψ where ψ is disjunctive. By the naturality of negation we have ϕ[f ] ≡ (¬ψ)[f ] ≡ ¬(ψ[f ]), so we may savely omit the parentheses. From the disjunctive case we learn that ψ[f ] is disjunctive. Ac- cording to the second part of Definition 3.1.13 this makes ϕ[f ] ≡ ¬ψ[f ] conjunctive, as required. In the same way we get

a ∈ ιX(ϕ) ⇔ a ∈ ιX(ψ) ⇔ Luf(a) ∈ ιY(ψ[f ]) ⇔ Luf(a) ∈ ιY(ϕ[f ]) and ϕ[f ]Lu f ≡ (¬ψ[f ])Luf(a)≡ ¬(ψ[f ]L u f(a)) ≡ ¬ψa[f ] ≡ ϕa[f ], just as needed. 

3.1. A FUNCTORIAL VERSION OF THE CONSTRUCTIBLE HIERARCHY 153

Our next goal is to show that the verification calculus is sound if X is isomorphic to an ordinal α. Proposition 3.1.3 interprets each term a ∈ Lu

X as a setJaK ∈ L

u αin

the actual constructible hierarchy. We can extend this interpretation to formulas: To transform the Lu_X-formula ϕ into an Luα-formula JϕK, replace any parameter a ∈ Lu_X in ϕ by the parameter_{JaK ∈ L}u_α. From Proposition 1.3.3 we know that the satisfaction relation Luα JϕK is primitive recursive.

Lemma 3.1.15. Assume that X is isomorphic to an ordinal α. Given a closed Lu_X-formula ϕ, if we have ϕ 'W a∈ιX(ϕ)ϕa (resp. ϕ ' V a∈ιX(ϕ)ϕa) then L u α JϕK is equivalent to ∃a∈ιX(ϕ)L u α JϕaK (resp. ∀a∈ιX(ϕ)L u αJϕaK).

Proof. One begins with the disjunctive clauses from Definition 3.1.13. As a first example, consider a formula

ϕ ≡ b ∈ {x ∈ Lu_s| θ(x, c)} 'W

suppL

X(a)<finXsa = b ∧ θ(a, c).

Let d be the Lu_X-term {x ∈ Lu_s| θ(x, c)}, so that Lu

α JϕK is equivalent to JbK ∈ JdK. Writing αs∈ α for the image of s under the isomorphism X ∼= α we must verify

JbK ∈ JdK ⇔ ∃a∈LuX(supp

L

X(a) <finX s ∧JaK = JbK ∧ L

u

α θ(JaK, JcK)).

Concerning direction ⇒, Proposition 3.1.3 does ensure that there is a term a ∈ Lu_X withJaK = JbK and supp

L

X(a) <finX suppLX(d). As we have suppLX(d) = {s}∪suppLX(c)

with suppL_X(c) <fin_X s the latter implies suppL_X(a) <fin_X s. Also,

JaK = JbK ∈ JdK = {x ∈ L

u αs| L

u

αs  θ(x,JcK)}

yields Luαs  θ(JaK, JcK). As θ is a ∆0-formula (cf. Definition 3.1.2) the missing

conjunct Luα  θ(JaK, JcK) follows from Lemma 1.3.6. As for direction ⇐, by Prop-

osition 3.1.3 the condition suppL_X(a) <fin_X s implies _{JaK ∈ L}u_αs. To deduce that JbK = JaK is an element of JdK it remains to establish L

u

αs  θ(JaK, JcK). The latter

follows from the assumption Lu

α θ(JaK, JcK), again by Lemma 1.3.6. Next, assume

that ϕ is of the form b0 6= b1. In this case the claim amounts to

Luα Jb0K 6= Jb1K ⇔ L

u

α ∃x∈Jb0Kx /∈Jb1K ∨ L u

α ∃x∈Jb1Kx /∈Jb0K.

This holds because the transitive set Luα (cf. Lemma 1.3.9) satisfies the axiom of

extensionality. As a last disjunctive case, consider a formula ϕ ≡ ∃xψ(x) 'W_a∈Lu

X ψ(a).

Here we have to verify

Luα  ∃xJψK(x) ⇔ ∃a∈LuXL

u

Concerning direction ⇒, Tarski’s conditions (see Proposition 1.3.3) yield a set x ∈ Lu

α with LuαJψK(x). By Proposition 3.1.3 the interpretation J·K : L

u

X → Luα is

surjective. Thus we get a term a ∈ Lu_X with Luα JψK(JaK). To conclude it suffices to observe that _{JψK(JaK) and Jψ(a)K are the same L}u_α-formula. In direction ⇐ one first infers Luα  JψK(JaK). By Proposition 3.1.3 we have JaK ∈ L

u α. Thus we get ∃_x∈Lu αL u α  JψK(x), and finally L u

α  ∃xJψK(x) by Tarski’s conditions. Having checked all disjunctive cases, let us now consider a conjunctive formula ϕ. As before we write ϕ ≡ ¬ψ where ψ is disjunctive, to get

ϕ 'V

a∈ιX(ψ)¬ψa.

Also observe_{J¬ψK ≡ ¬JψK. Using the claim for the disjunctive case we obtain} Luα JϕK ⇔ L u α 2 JψK ⇔ ∀a∈ιX(ψ)L u α2 JψaK ⇔ ∀a∈ιX(ψ)L u α J¬ ψaK, just as required. 

To make use of soundness we will need to know that the relevant verifications are well-founded. As a first step, let us make the verification order explicit. Func- toriality of this order will not be required.

Definition 3.1.16. The order <ιon closed LuX-formulas is defined as the trans-

itive closure of the relation

ϕa<ι ϕ for each a ∈ ιX(ϕ).

Note that the transitive closure of a binary relation can be constructed in prim- itive recursive set theory, similar to the proof of Proposition 1.1.11. As promised we have the following:

Proposition 3.1.17. If X is isomorphic to an ordinal then the verification relation <ι on Lu_X-formulas is well-founded.

We could prove the same result for any well-order X, whether or not we have an isomorphism with an ordinal. However, such an isomorphism will be given in our application and it makes the presentation of the proof easier.

Proof. To accomodate the elements of u we shift the given isomorphism by one, so that we get an order embedding X 3 s 7→ αs∈ α with αs> 0 for all s ∈ X.

Now we define a notion of ordinal height for Lu_X-terms, setting

3.1. A FUNCTORIAL VERSION OF THE CONSTRUCTIBLE HIERARCHY 155

Some readers may have expected the height of {x ∈ Lu_s| ϕ(x, ~a)} to depend on the formula ϕ(x, ~a). In our set-up it does not, because we have required ϕ(x, ~a) to be a ∆0-formula with parameters “below s” (cf. Definition 3.1.2). Following the proof-

theoretic literature (see in particular [38]) we extend the notion of ordinal height to closed Lu_X-formulas. Assumption 3.1.1 ensures that the parameter 0 ∈ u ⊆ Lu_X is available for any order X. By recursion over the length of formulas we define

ht(a ∈ b) := ht(a /∈ b) := max{ht(a) + 6, ht(b) + 1}, ht(a = b) := ht(a 6= b) := max{ht(a), ht(b), 5} + 4, ht(ϕ0∨ ϕ1) := ht(ϕ0∧ ϕ1) := max{ht(ϕ0), ht(ϕ1)} + 1,

ht(∃x∈bϕ(x)) := ht(∀x∈bϕ(x)) := max{ht(b), ht(ϕ(0)) + 2},

ht(∃xϕ(x)) := ht(∀xϕ(x)) := max{ω · α, ht(ϕ(0)) + 1}.

To establish the proposition it suffices to show

ht(ϕa) < ht(ϕ) for any LuX-formula ϕ and any a ∈ ιX(ϕ).

As a preparation, observe ht(a) < ω · αs for any a ∈ LuX with suppLX(a) <finX s.

Note that this implies ht(a) + n < ω · αs for any n ∈ N, because ω · αs is a limit

ordinal. If ϕ is a ∆0-formula such that suppL_X(a) <fin_X s holds for any parameter a

in ϕ, then we get ht(ϕ) < ω · αs. For a general Lu_X-formula ϕ ≡ ϕ(x) one shows

ht(ϕ(a)) < max{ω · αs, ht(ϕ(0)) + 1} for a ∈ LuX with suppLX(a) <finX s,

by induction over the length of ϕ (cf. [38, Lemma 3]). Similarly one sees ht(ϕ(a)) < max{ω · α, ht(ϕ(0)) + 1} for all a ∈ Lu_X.

Based on these facts it is straightforward to verify ht(ϕa) < ht(ϕ) for any disjunct-

ive formula ϕ. As a representative example, let us consider ϕ ≡ ∃_x∈{y∈Lu s| θ(y,c)}ψ(x) ' W suppL X(a)< fin Xsθ(a, c) ∧ ψ(a).

Observe that we have

ht(ϕ) = max{ω · αs, ht(ψ(0)) + 2}

and

ht(ϕa) = ht(θ(a, c) ∧ ψ(a)) = max{ht(θ(a, c)) + 1, ht(ψ(a)) + 1}.

By the definition of Lu_X-terms we have suppL_X(c) <fin_X s, so that we get ht(θ(a, c)) + 1 < ω · αs ≤ ht(ϕ).

We also have

ht(ψ(a)) + 1 < max{ω · αs, ht(ϕ(0)) + 2} = ht(ϕ),

as required. Having checked all disjunctive cases, consider a conjunctive formula ϕ ≡ ¬ψ where ψ is disjunctive. According to Definition 3.1.13 we have

ϕ 'V

a∈ιX(ψ)¬ ψa.

By the disjunctive case we obtain

ht(ϕa) = ht(¬ ψa) = ht(ψa) < ht(ψ) = ht(¬ψ) = ht(ϕ)

for any a ∈ ιX(ψ) = ιX(ϕ), just as needed. 

3.2. From Search Tree to Admissible Set

In the introduction to this chapter we have sketched how Sch¨utte’s method of search trees (deduction chains) can be used to construct well-founded models of set theories. The details of this construction are worked out in the present section.

Our aim is to construct a search tree S_Xu for any linear order X and any set u which satisfies Assumption 3.1.1. Recall the syntactic version Lu_X of the constructible hierarchy that was defined in the previous section. According to Proposition 1.2.8 we can form the set (Lu_X)<ωof finite sequences with entries in Lu_X. As usual we obtain a tree if we order these sequences by end extension. The search tree S_Xu will be defined as a labelled subtree of (Lu_X)<ω.

Let us discuss some terminology that will be needed in the definition of S_Xu: First, recall that we write len(σ) for the length of a sequence σ = ha0, . . . , alen(σ)−1i.

If n ≤ len(σ) then σ  n denotes the restriction of σ to its first n entries, i.e. we have σ  n = ha0, . . . , an−1i. In the previous section we have considered support

functions suppL

X : LuX → [X]<ω. To get functions suppXS : (LuX)<ω → [X]<ω we set

suppS_X(ha0, . . . , an−1i) := suppXL(a0) ∪ · · · ∪ suppLX(an−1).

Note that the family of functions suppS_· is primitive recursive by Corollary 1.2.11. In the next section we will see that the functions suppS_X compute the support of a certain prae-dilator.

Next, recall the notion of Lu_X-formula from Definition 3.1.12. By an Lu_X-sequent we mean a finite sequence Γ = hϕ0, . . . , ϕlen(Γ)−1i of closed LuX-formulas ϕi. Intu-

itively Γ should be interpreted as the disjunction ϕ0∨ · · · ∨ ϕlen(Γ)−1. In particular

In document Follow this and additional works at: (página 64-67)