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Computability and Logic

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282 SECOND-ORDER LOGIC<br />

is true. But N0 is true, since the individual a that is the denotation of 0 is in the set<br />

A that is the denotation of N, <strong>and</strong> ∀x(Nx → Nx ′ ) is true, since if any individual is in<br />

A, so is the value obtained when the function f that is the denotation of ′ is applied<br />

to that individual as argument. Hence ∀xNx must be true, <strong>and</strong> this means that every<br />

individual in the domain is in A, so the domain, being just A, is enumerable.<br />

Conversely, suppose that the domain of an interpretation M is enumerable. Fix<br />

an enumeration of its elements: m 0 , m 1 , m 2 , <strong>and</strong> so on. Let a be m 0 , <strong>and</strong> let f be the<br />

function that given m i as argument yields m i+1 as value, <strong>and</strong> add a constant 0 <strong>and</strong> a<br />

one-place function symbol ′ to denote a <strong>and</strong> f . Given any subset A of the domain,<br />

suppose we add a one-place predicate N to denote A. Then if N0 is true, a = m 0<br />

must belong to A, <strong>and</strong> if ∀x(Nx → Nx ′ ) is true, then whenever m i belongs to A,<br />

f (m i ) = m i+1 must belong to A. So if both are true, every element m 0 , m 1 , m 2 ,...<br />

of the domain must belong to A, <strong>and</strong> therefore ∀xNx is true. Thus<br />

(N0 & ∀x(Nx → Nx ′ )) →∀xNx<br />

is true if N is taken to denote A, <strong>and</strong> therefore A satisfies<br />

<strong>and</strong> since this is true for any A,<br />

is true, <strong>and</strong> therefore<br />

or Enum is true in M.<br />

(X0 & ∀x(Xx → Xx ′ )) →∀xXx<br />

∀X((X0 & ∀x(Xx → Xx ′ )) →∀xXx)<br />

∃z∃u∀X((Xz& ∀x(Xx → Xu(x))) →∀xXx)<br />

22.4 Example (The ‘axiom’ of infinity). Let Inf be the sentence<br />

∃z∃u(∀xz≠ u(x)&∀x∀y(u(x) = u(y) → x = y)).<br />

Then Inf is true in an interpretation if <strong>and</strong> only if its domain is infinite. The proof is left to<br />

the reader.<br />

22.5 Proposition. The downward <strong>and</strong> upward Löwenheim–Skolem theorems both fail<br />

for second-order logic.<br />

Proof: Inf & ∼Enum <strong>and</strong> Inf & Enum are both second-order sentences having<br />

infinite but no finite models. The former has only nonenumerable models, contrary<br />

to the downward Löwenheim–Skolem theorem; the latter only denumerable models,<br />

contrary to the upward Löwenheim–Skolem theorem.<br />

It is an immediate consequence of the downward <strong>and</strong> upward Löwenheim–Skolem<br />

theorems that if a first-order sentence or set of such sentences has an infinite model,<br />

then it has nonisomorphic infinite models. Even this corollary of the Löwenheim–<br />

Skolem theorems fails for second-order logic, as the next example shows.

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