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Math 430: Problem Set VI due: Friday, 4/
Proposition 0.1 (Inference). Let Γ ⊆ L and ϕ, ψ ∈ L. If Γ ϕ and Γ (ϕ → ψ) then Γ ` ψ.
We make use of the preceding Proposition freely (the proof is similar to Lemma 1.26.)
Exercises on L proof system:
(1) Provide a proof of the Deduction lemma for L (Theorem 5.9): If Γ ∪ {ϕ} ψ then Γ (ϕ → ψ). Use Lemma 1.28 as your model, but with the following simplification. Assume that the (IH) is true for all proofs of length n, then prove it true for an arbitrary proof of length n + 1: (IH) For all m < n, If 〈ψ 1 ,... , ψm〉 is a Γ ∪ {ϕ}-proof, then Γ (ϕ → ψm). The base case n = 1 is trivial, as there are no proofs of length m < n. Notice that your (IH) establishes that in fact, for a Γ ∪ {ϕ}-proof 〈ψ 1 ,... , ψn〉, Γ (ϕ → ψl) for all l < (n − 1).
(2) (Theorem 5.10, Generalization) Suppose xi does not occur free in any ψ ∈ Γ. Prove that if Γ ϕ then Γ (∀xiϕ). Here is an outline of a proof: Assume that xi does not occur free in any ψ ∈ Γ. Assume that for any formula ϕ if Γ proves ϕ with a proof of length m < n, then Γ (∀xiϕ). Fix a formula ϕ such that Γ proves ϕ with a proof of length n. (Show that Γ (∀xiϕ).) Let 〈θ 1 ,... , θn〉 be a proof of this ϕ from Γ (so θn = ϕ.) There are three cases to consider,
- θn ∈ Γ, 2. θn ∈ ∆, 3. θn is obtained by modus ponens from previous θi, θj = (θi → θn). In any of these cases, you need to show that Γ ` (∀xiθn). Hints for case 1. and case 3. can be found in the text in the proof of Theorem 5.10. What about case 2.?
As in the propositional case, by a Completeness theorem for our logic, we may deduce a Com- pactness theorem.
Theorem 0.2 (Compactness). If Γ ⊆ L and every finite Γ 0 ⊆ Γ is satisfiable, then Γ is satisfiable.
Notation 0.3. In the case that T ⊆ LA is a set of sentences, we call T a theory in LA.
Recall that for a set Γ ⊆ L we say that (M, ν) Γ if (M, ν) ψ for all ψ ∈ Γ. We have a special notation for the case of theories.
Notation 0.4. (1) Given a theory T in LA and an interpretation M T we refer to M as a model of T. 1
(2) For a finite set X, by |X| we refer to the size of X. (3) We say that a structure M is infinite in the case that the underlying set, M , is infinite. Below we define the theory of symmetric graphs with no loops, but we call them graphs:
Definition 0.5. Let A = {R} for a binary predicate symbol R. Define the theory of graphs Tg to be the following set of LA-sentences:
{(∀x(¬R(x, x)), (∀x(∀y (R(x, y) → R(y, x)) ))}
Definition 0.6. Let K be a class of LA-structures. We say that K is an elementary class if there is some theory T in LA such that for any LA-structure M:
M T ⇔ M ∈ K
Exercises on Compactness:
(3) One version of G¨odel Completeness states that a set Γ ⊆ L is consistent if and only if it is satisfiable. Use this version to prove the Compactness theorem, above. You may make use of Lemma 1.30 for the case of L-formulas (the proof is similar.) In the first-order logic case we have a designated formula ψ := (¬x 1 =ˆx 1 ) that is false in all interpretations. So in effect, this is a concrete formula that may play the role of (A 0 ∧ (¬A 0 )) in this new context.
(4) Suppose T is a theory in LA and {cs : s ∈ S} is a set of constants not occuring in T (so S ⊆ N.) Suppose that for any n ∈ N, T has a model M such that |M | ≥ n. Show that T has an infinite model.
(5) Let A = {R, c} for a binary predicate R and let T be an LA-theory extending Tg. (Thus every model of T is a symmetric graph with no loops that interprets c as a specific vertex in the domain.) Suppose that for every n ∈ N, there is a model Mn T such that cMn^ is RMn^ -connected to at least n distinct vertices. Show that there is a model of T where c is interpreted to be connected to infinitely many vertices.
(6) Show that the class of LA-structures which are graphs and interpret c to be connected to finitely many points is not an elementary class. [For contradiction, assume such a theory T exists, you may as well assume that T ⊇ Tg. Now use the previous exercise.]
