2 attempt to clarify further edited May 20 '14 at 12:45 reinierpost 3,3871616 silver badges2727 bronze badges The theorem says that when a sentence has arbitrarily large (finite) models, then it also has infinite models. The antecedent of the theorem: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a model of $$\phi$$ with at least $$n$$ elements. doesn'tis equivalent to saying that there are infinitely many different models, but it does not state that any of thethese models are infinite in size. In fact, it would be clearer to use: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a finite model of $$\phi$$ with at least $$n$$ elements. This is not exactly the same statement, but as an antecedent within this theorem, it is equivalent. The consequent: $$\phi$$ has a model with infinitely many elements. doesn'tdoes not say anything about how many different models there are - it just says something about the size of at least one of them. To see that the truth of this theorem doesn't simply follow from its structure, consider the following statement, which has the same structure: Let $$\phi$$ be a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is an equivalent sentence of predicate logic with at least $$n$$ elements. Then there is an equivalent sentence with infinitely many elements. Is this a theorem, too? No, this statement is false, for the simple reason that all sentences of predicate logic have finitely many elements by definition. The theorem says that when a sentence has arbitrarily large (finite) models, it also has infinite models. The antecedent of the theorem: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a model of $$\phi$$ with at least $$n$$ elements. doesn't state that any of the models are infinite in size. In fact, it would be clearer to use: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a finite model of $$\phi$$ with at least $$n$$ elements. This is not exactly the same statement, but as an antecedent within this theorem, it is equivalent. The consequent: $$\phi$$ has a model with infinitely many elements. doesn't say anything about how many different models there are - it just says something about the size of at least one of them. To see that the truth of this theorem doesn't simply follow from its structure, consider the following statement, which has the same structure: Let $$\phi$$ be a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is an equivalent sentence of predicate logic with at least $$n$$ elements. Then there is an equivalent sentence with infinitely many elements. Is this a theorem, too? No, this statement is false, for the simple reason that all sentences of predicate logic have finitely many elements by definition. The theorem says that when a sentence has arbitrarily large (finite) models, then it also has infinite models. The antecedent of the theorem: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a model of $$\phi$$ with at least $$n$$ elements. is equivalent to saying that there are infinitely many different models, but it does not state that any of these models are infinite in size. In fact, it would be clearer to use: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a finite model of $$\phi$$ with at least $$n$$ elements. This is not exactly the same statement, but as an antecedent within this theorem, it is equivalent. The consequent: $$\phi$$ has a model with infinitely many elements. does not say anything about how many different models there are - it just says something about the size of at least one of them. To see that the truth of this theorem doesn't simply follow from its structure, consider the following statement, which has the same structure: Let $$\phi$$ be a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is an equivalent sentence of predicate logic with at least $$n$$ elements. Then there is an equivalent sentence with infinitely many elements. Is this a theorem, too? No, this statement is false, for the simple reason that all sentences of predicate logic have finitely many elements by definition. 1 answered Apr 21 '14 at 13:38 reinierpost 3,3871616 silver badges2727 bronze badges The theorem says that when a sentence has arbitrarily large (finite) models, it also has infinite models. The antecedent of the theorem: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a model of $$\phi$$ with at least $$n$$ elements. doesn't state that any of the models are infinite in size. In fact, it would be clearer to use: $$\phi$$ is a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is a finite model of $$\phi$$ with at least $$n$$ elements. This is not exactly the same statement, but as an antecedent within this theorem, it is equivalent. The consequent: $$\phi$$ has a model with infinitely many elements. doesn't say anything about how many different models there are - it just says something about the size of at least one of them. To see that the truth of this theorem doesn't simply follow from its structure, consider the following statement, which has the same structure: Let $$\phi$$ be a sentence of predicate logic such that for any natural number $$n \geq 1$$, there is an equivalent sentence of predicate logic with at least $$n$$ elements. Then there is an equivalent sentence with infinitely many elements. Is this a theorem, too? No, this statement is false, for the simple reason that all sentences of predicate logic have finitely many elements by definition.