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The answer is yes, provided these theories are consistent. For example, PA proves that there is a number $n$, such that if there is any proof of a contradiction in PA, then there is one of size less than $n$. This is trivial if you think about it, since either there is a proof of a contradiction in PA, in which case you can use $n$ to be any number larger than it, or else there isn't, in which case you can use any $n$. But PA does not prove that any particular number $n$ has this property, since then PA would prove its own consistency, which violates the incompleteness theorem.

Here is another kind of example: Let $\sigma$ be any statement that is independent of PA, and let $\phi(n)$ assert that $(n=0\wedge\sigma)\vee(n=1\wedge\neg\sigma)$. So PA proves $\phi(0)\vee \phi(1)$, and in particular it proves $\exists x\ \phi(x)$, but PA doesn't prove either case, because $\sigma$ is independent.

If you strengthen the conclusion, however, to say that the theory should actually prove $\neg\phi(n)$ for every numeral $n$, then you would be asking whether these theories are $\omega$-inconsistent. Of course, we all think that our theories are $\omega$-consistent, because we want and expect that the natural numbers referred to inside the theory are somehow the same as those in the meta theory. But of course, because of the incompleteness theorem, we cannot prove this within the theory itself. Notice that a theory is $\omega$-consistent, if it has an $\omega$-model, a model in which the natural numbers of the model are the same as the natural numbers of the meta-theory. For this reason, our stronger theories can often prove that our weaker theories are $\omega$-consistent. For example, ZFC proves that PA is $\omega$-consistent, and ZFC+large cardinals proves that ZFC is $\omega$-consistent. My own view is the assertion that a theory is $\omega$-consistent is just one stop along a continuum, from assuming consistency up to iterated consistency up to having a well-founded model up to large cardinals and so on, moving up in consistency strength at each step.

Meanwhile, if these theories are consistent, then it is easy to make examples of $\omega$-inconsistent theories extending them. For example, the theories PA+$\neg$Con(PA) and ZFC+$\neg$Con(ZFC) are consistent by the incompleteness theorem, but not $\omega$-consistent, because each of them asserts that there is a certain proof of a contradiction, which is provably not instantiated by any actual natural number witness.

The answer is yes, provided these theories are consistent. For example, PA proves that there is a number $n$, such that if there is no proof of a contradiction in PA of size at most $n$, then there is no proof of a contradiction in PA at all. This is trivial if you think about it, since either there is a proof of a contradiction in PA, in which case you can let $n$ be any number larger than that proof, or else there isn't such a proof of a contradiction, in which case any $n$ has the desired property. But PA does not prove that any particular number $n$ has this property, since then PA would prove its own consistency, which violates the incompleteness theorem.

Here is another kind of example: Let $\sigma$ be any statement that is independent of PA, and let $\phi(n)$ assert that $(n=0\wedge\sigma)\vee(n=1\wedge\neg\sigma)$. So PA proves $\phi(0)\vee \phi(1)$, and in particular it proves $\exists x\ \phi(x)$, but PA doesn't prove either case separately, because $\sigma$ is independent.

If you strengthen the conclusion of your question, however, to say that the theory should actually prove $\neg\phi(n)$ for every numeral $n$, then you would be asking whether these theories are $\omega$-inconsistent. Of course, we all think that our theories are $\omega$-consistent, because we want and expect that the natural numbers referred to inside the theory are somehow the same as those in the meta theory. But of course, because of the incompleteness theorem, we cannot prove this within the theory itself. Notice that a theory must be $\omega$-consistent in the case that it has an $\omega$-model, a model in which the natural numbers of the model are the same as the natural numbers of the meta-theory. For this reason, our stronger theories can often prove that our weaker theories are $\omega$-consistent. For example, ZFC proves that PA is $\omega$-consistent, and ZFC+large cardinals proves that ZFC is $\omega$-consistent. My own view is the assertion that a theory is $\omega$-consistent is just one stop along a continuum, from assuming consistency up to iterated consistency up to having a well-founded model up to large cardinals and so on, moving up in consistency strength at each step.

Meanwhile, if these theories are consistent, then it is easy to make examples of $\omega$-inconsistent theories extending them. For example, the theories PA+$\neg$Con(PA) and ZFC+$\neg$Con(ZFC) are consistent by the incompleteness theorem, but not $\omega$-consistent, because each of them asserts that there is a certain proof of a contradiction, which is provably not instantiated by any actual natural number witness.

The answer is yes, provided these theories are consistent. For example, PA proves that there is a number $n$, such that if there is any proof of a contradiction in PA, then there is one of size less than $n$. This is trivial if you think about it, since either there is a proof of a contradiction in PA, in which case you can use $n$ to be any number larger than it, or else there isn't, in which case you can use any $n$. But PA does not prove that any particular number $n$ has this property, since then PA would prove its own consistency, which violates the incompleteness theorem.

Here is another kind of example: Let $\sigma$ be any statement that is independent of PA, and let $\phi(n)$ assert that $(n=0\wedge\sigma)\vee(n=1\wedge\neg\sigma)$. So PA proves $\phi(0)\vee \phi(1)$, and in particular it proves $\exists x\ \phi(x)$, but PA doesn't prove either case, because $\sigma$ is independent.

If you strengthen the conclusion, however, to say that the theory should actually prove $\neg\phi(n)$ for every numeral $n$, then you would be asking whether these theories are $\omega$-inconsistent. Of course, we all think that our theories are $\omega$-consistent, because we want and expect that the natural numbers referred to inside the theory are somehow the same as those in the meta theory. But of course, because of the incompleteness theorem, we cannot prove this within the theory itself. A theory is $\omega$-consistent just in case it has an $\omega$-model, a model in which the natural numbers of the model are the same as the natural numbers of the meta-theory.

Nevertheless, our stronger theories can often prove that our weaker theories are $\omega$-consistent. For example, ZFC proves that PA is $\omega$-consistent, and ZFC+large cardinals proves that ZFC is $\omega$-consistent. My own view is the assertion that a theory is $\omega$-consistent is just one stop along a continuum, from assuming consistency up to iterated consistency up to having a well-founded model up to large cardinals and so on, moving up in consistency strength at each step.

Meanwhile, if these theories are consistent, then it is easy to make examples of $\omega$-inconsistent theories extending them. For example, the theories PA+$\neg$Con(PA) and ZFC+$\neg$Con(ZFC) are consistent by the incompleteness theorem, but not $\omega$-consistent, because each of them asserts that there is a certain proof of a contradiction, which cannot be instantiated by any actual natural number witness.

The answer is yes, provided these theories are consistent. For example, PA proves that there is a number $n$, such that if there is any proof of a contradiction in PA, then there is one of size less than $n$. This is trivial if you think about it, since either there is a proof of a contradiction in PA, in which case you can use $n$ to be any number larger than it, or else there isn't, in which case you can use any $n$. But PA does not prove that any particular number $n$ has this property, since then PA would prove its own consistency, which violates the incompleteness theorem.

Here is another kind of example: Let $\sigma$ be any statement that is independent of PA, and let $\phi(n)$ assert that $(n=0\wedge\sigma)\vee(n=1\wedge\neg\sigma)$. So PA proves $\phi(0)\vee \phi(1)$, and in particular it proves $\exists x\ \phi(x)$, but PA doesn't prove either case, because $\sigma$ is independent.

If you strengthen the conclusion, however, to say that the theory should actually prove $\neg\phi(n)$ for every numeral $n$, then you would be asking whether these theories are $\omega$-inconsistent. Of course, we all think that our theories are $\omega$-consistent, because we want and expect that the natural numbers referred to inside the theory are somehow the same as those in the meta theory. But of course, because of the incompleteness theorem, we cannot prove this within the theory itself. Notice that a theory is $\omega$-consistent, if it has an $\omega$-model, a model in which the natural numbers of the model are the same as the natural numbers of the meta-theory. For this reason, our stronger theories can often prove that our weaker theories are $\omega$-consistent. For example, ZFC proves that PA is $\omega$-consistent, and ZFC+large cardinals proves that ZFC is $\omega$-consistent. My own view is the assertion that a theory is $\omega$-consistent is just one stop along a continuum, from assuming consistency up to iterated consistency up to having a well-founded model up to large cardinals and so on, moving up in consistency strength at each step.

Meanwhile, if these theories are consistent, then it is easy to make examples of $\omega$-inconsistent theories extending them. For example, the theories PA+$\neg$Con(PA) and ZFC+$\neg$Con(ZFC) are consistent by the incompleteness theorem, but not $\omega$-consistent, because each of them asserts that there is a certain proof of a contradiction, which is provably not instantiated by any actual natural number witness.

The answer is yes, provided these theories are consistent. For example, PA proves that there is a number $n$, such that if there is any proof of a contradiction in PA, then there is one of size less than $n$. This is trivial if you think about it, since either there is a proof of a contradiction in PA, in which case you can use $n$ to be any number larger than it, or else there isn't, in which case you can use any $n$. But PA does not prove that any particular number $n$ has this property, since then PA would prove its own consistency, which violates the incompleteness theorem.

Here is another kind of example: Let $\sigma$ be any statement that is not provable in PA, and let $\phi(n)$ assert that $(n=0\wedge\sigma)\vee(n=1\wedge\neg\sigma)$. So PA proves either $\phi(0)$ or $\phi(1)$, but PA doesn't prove either case, if consistent, because $\sigma$ is independent.

If you strengthen the conclusion, however, to say that the theory should actually prove $\neg\phi(n)$ for every numeral $n$, then you would be asking whether these theories are $\omega$-inconsistent. Of course, we all think that our theories are $\omega$-consistent, because we want and expect that the natural numbers referred to inside the theory are somehow the same as those in the meta theory. But of course, because of the incompleteness theorem, we cannot prove this within the theory itself. A theory is $\omega$-consistent just in case it has an $\omega$-model, a model in which the natural numbers of the model are the same as the natural numbers of the meta-theory.

Nevertheless, our stronger theories can often prove that our weaker theories are $\omega$-consistent. For example, ZFC proves that PA is $\omega$-consistent, and ZFC+large cardinals proves that ZFC is $\omega$-consistent. My own view is the assertion that a theory is $\omega$-consistent is just one stop along a continuum, from assuming consistency up to iterated consistency up to having a well-founded model up to large cardinals and so on, moving up in consistency strength at each step.

Meanwhile, if these theories are consistent, then it is easy to make examples of $\omega$-inconsistent theories extending them. For example, the theories PA+$\neg$Con(PA) and ZFC+$\neg$Con(ZFC) are consistent by the incompleteness theorem, but not $\omega$-consistent, because each of them asserts that there is a certain proof of a contradiction, which cannot be instantiated by any actual natural number witness.

The answer is yes, provided these theories are consistent. For example, PA proves that there is a number $n$, such that if there is any proof of a contradiction in PA, then there is one of size less than $n$. This is trivial if you think about it, since either there is a proof of a contradiction in PA, in which case you can use $n$ to be any number larger than it, or else there isn't, in which case you can use any $n$. But PA does not prove that any particular number $n$ has this property, since then PA would prove its own consistency, which violates the incompleteness theorem.

Here is another kind of example: Let $\sigma$ be any statement that is independent of PA, and let $\phi(n)$ assert that $(n=0\wedge\sigma)\vee(n=1\wedge\neg\sigma)$. So PA proves $\phi(0)\vee \phi(1)$, and in particular it proves $\exists x\ \phi(x)$, but PA doesn't prove either case, because $\sigma$ is independent.

If you strengthen the conclusion, however, to say that the theory should actually prove $\neg\phi(n)$ for every numeral $n$, then you would be asking whether these theories are $\omega$-inconsistent. Of course, we all think that our theories are $\omega$-consistent, because we want and expect that the natural numbers referred to inside the theory are somehow the same as those in the meta theory. But of course, because of the incompleteness theorem, we cannot prove this within the theory itself. A theory is $\omega$-consistent just in case it has an $\omega$-model, a model in which the natural numbers of the model are the same as the natural numbers of the meta-theory.

Nevertheless, our stronger theories can often prove that our weaker theories are $\omega$-consistent. For example, ZFC proves that PA is $\omega$-consistent, and ZFC+large cardinals proves that ZFC is $\omega$-consistent. My own view is the assertion that a theory is $\omega$-consistent is just one stop along a continuum, from assuming consistency up to iterated consistency up to having a well-founded model up to large cardinals and so on, moving up in consistency strength at each step.

Meanwhile, if these theories are consistent, then it is easy to make examples of $\omega$-inconsistent theories extending them. For example, the theories PA+$\neg$Con(PA) and ZFC+$\neg$Con(ZFC) are consistent by the incompleteness theorem, but not $\omega$-consistent, because each of them asserts that there is a certain proof of a contradiction, which cannot be instantiated by any actual natural number witness.