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As is being discussed in the comments, the non-standard proofs in non-standard models of $\mathsf{PA}$ are not trustworthy (in that they're not sound). I'll give a sort of 'toy model' illustrating the intuition about why this is the case.

In a Hilbert-style deduction calculus you have some starting set of axioms and one or more deduction rules (often just modus ponens). A proof is a sequence $$ \varphi_0,\varphi_1,\dots,\varphi_n$$ of sentences satisfying that for each $i\leq n$, either $\varphi_i$ is an axiom or there are $j,k<i$ such that $\varphi_k = (\varphi_j \to \varphi_i)$.

Given such a derivation, the argument that we should believe $\varphi_n$ (provided that we believe in the axioms and in modus ponens) is ultimately an inductive proof. We argue that we should believe $\varphi_0$ and that if we believe $\varphi_i$ for all $i<j$, then we should believe $\varphi_j$.

Non-standard models are a lot more complicated than what I'm about to describe, but this really gets to the heart of the matter. Suppose we're in a non-standard model of arithmetic and the numbers look like this: $$ 0, 1, 2, \dots \infty - 2, \infty - 1, \infty, \infty + 1, \infty +2,\dots,$$ where '$\infty$' is some fixed infinite non-standard number. Now suppose I give you a derivation that lives in this model which looks like this: $$\varphi_0, \varphi_1,\varphi_{-2},\dots, \varphi_{\infty - 2},\varphi_{\infty-1}, \varphi_{\infty}.$$ This may very well be a derivation in the above sense (as in each step is either an axiom or follows from two previous steps by modus ponens), but the existence of such an object doesn't imply that $\varphi_\infty$ is a trustworthy statement. In fact (and this is the really toy part as I'll explain in a second), we could just have

• $\varphi_i = ((0= 1)\to(0 = 1))$ for each (externally) finite $i$,
• $\varphi_{\infty+i} = (0 = 1)$ for each integer $i$.

This sequence of sentences satisfies the definition (each $\varphi_i$ for $i$ finite is an axiom and $\varphi_{\infty+i}$ follows from $\varphi_0$ and $\varphi_{\infty + i -1}$ by modus ponens) but clearly the conclusion is not supported because the proof of it is relying on an infinite regress. (And this is even ignoring the possibility of sentences that are infinitely long, which is another issue.)

Now, the reason that this is a toy model is that clearly we have a failure of induction. The sequence I've defined satisfies that $\varphi_0 = (0\neq 1)$ and that for each $i$, if $\varphi_i = (0 \neq 1)$, then $\varphi_{i+1} = (0\neq 1)$, yet it is not true that for all $i$, $\varphi_i = (0\neq 1)$. In fact, non-standard models of arithmetic are much harder to describe than what I've given above (and in some sense cannot be described 'explicitly'). In an actual model of $\mathsf{PA} + \neg\mathrm{Con}(\mathsf{PA})$, the proof of $0=1$ is ultimately unsound for the same reason (ill-foundedness of the implicit proof tree, or in other words a conclusion relying on an infinite regress), but the 'gap' or 'jump' is much harder to locate in some precise sense.

As is being discussed in the comments, the non-standard proofs in non-standard models of $\mathsf{PA}$ are not trustworthy (in that they're not sound). I'll give a sort of 'toy model' illustrating the intuition about why this is the case.

In a Hilbert-style deduction calculus you have some starting set of axioms and one or more deduction rules (often just modus ponens). A proof is a sequence $$ \varphi_0,\varphi_1,\dots,\varphi_n$$ of sentences satisfying that for each $i\leq n$, either $\varphi_i$ is an axiom or there are $j,k<i$ such that $\varphi_k = (\varphi_j \to \varphi_i)$.

Given such a derivation, the argument that we should believe $\varphi_n$ (provided that we believe in the axioms and in modus ponens) is ultimately an inductive proof. We argue that we should believe $\varphi_0$ and that if we believe $\varphi_i$ for all $i<j$, then we should believe $\varphi_j$.

Non-standard models are a lot more complicated than what I'm about to describe, but this really gets to the heart of the matter. Suppose we're in a non-standard model of arithmetic and the numbers look like this: $$ 0, 1, 2, \dots, \infty - 2, \infty - 1, \infty, \infty + 1, \infty +2,\dots,$$ where '$\infty$' is some fixed infinite non-standard number. Now suppose I give you a derivation that lives in this model which looks like this: $$\varphi_0, \varphi_1,\varphi_{-2},\dots, \varphi_{\infty - 2},\varphi_{\infty-1}, \varphi_{\infty}.$$ This may very well be a derivation in the above sense (as in each step is either an axiom or follows from two previous steps by modus ponens), but the existence of such an object doesn't imply that $\varphi_\infty$ is a trustworthy statement. In fact (and this is the really toy part as I'll explain in a second), we could just have

• $\varphi_i = ((0= 1)\to(0 = 1))$ for each (externally) finite $i$,
• $\varphi_{\infty+i} = (0 = 1)$ for each integer $i$.

This sequence of sentences satisfies the definition (each $\varphi_i$ for $i$ finite is an axiom and $\varphi_{\infty+i}$ follows from $\varphi_0$ and $\varphi_{\infty + i -1}$ by modus ponens) but clearly the conclusion is not supported because the proof of it is relying on an infinite regress. (And this is even ignoring the possibility of sentences that are infinitely long, which is another issue.)

Now, the reason that this is a toy model is that clearly we have a failure of induction. The sequence I've defined satisfies that $\varphi_0 = \varphi_0$ and that for each $i$, if $\varphi_i = \varphi_0$, then $\varphi_{i+1} = \varphi_0$, yet it is not true that for all $i$, $\varphi_i = \varphi_0$. In fact, non-standard models of arithmetic are much harder to describe than what I've given above (and in some sense cannot be described 'explicitly'). In an actual model of $\mathsf{PA} + \neg\mathrm{Con}(\mathsf{PA})$, the proof of $0=1$ is ultimately unsound for the same reason (ill-foundedness of the implicit proof tree, or in other words a conclusion relying on an infinite regress), but the 'gap' or 'jump' is much harder to locate in some precise sense.

As is being discussed in the comments, the non-standard proofs in non-standard models of $\mathsf{PA}$ are not trustworthy (in that they're not sound). I'll give a sort of 'toy model' illustrating the intuition about why this is the case.

In a Hilbert-style deduction calculus you have some starting set of axioms and one or more deduction rules (often just modus ponens). A proof is a sequence $$ \varphi_0,\varphi_1,\dots,\varphi_n$$ of sentences satisfying that for each $i\leq n$, either $\varphi_i$ is an axiom or there are $j,k<i$ such that $\varphi_k = (\varphi_j \to \varphi_j)$.

Given such a derivation, the argument that we should believe $\varphi_n$ (provided that we believe in the axioms and in modus ponens) is ultimately an inductive proof. We argue that we should believe $\varphi_0$ and that if we believe $\varphi_i$ for all $i<j$, then we should believe $\varphi_j$.

Non-standard models are a lot more complicated than what I'm about to describe, but this really gets to the heart of the matter. Suppose we're in a non-standard model of arithmetic and the numbers look like this: $$ 0, 1, 2, \dots \infty - 2, \infty - 1, \infty, \infty + 1, \infty +2,\dots,$$ where '$\infty$' is some fixed infinite non-standard number. Now suppose I give you a derivation that lives in this model which looks like this: $$\varphi_0, \varphi_1,\varphi_{-2},\dots, \varphi_{\infty - 2},\varphi_{\infty-1}, \varphi_{\infty}.$$ This may very well be a derivation in the above sense (as in each step is either an axiom or follows from two previous steps by modus ponens), but the existence of such an object doesn't imply that $\varphi_\infty$ is a trustworthy statement. In fact (and this is the really toy part as I'll explain in a second), we could just have

• $\varphi_i = ((0= 1)\to(0 = 1))$ for each (externally) finite $i$,
• $\varphi_{\infty+i} = (0 = 1)$ for each integer $i$.

This sequence of sentences satisfies the definition (each $\varphi_i$ for $i$ finite is an axiom and $\varphi_{\infty+i}$ follows from $\varphi_0$ and $\varphi_{\infty + i -1}$ by modus ponens) but clearly the conclusion is not supported because the proof of it is relying on an infinite regress. (And this is even ignoring the possibility of sentences that are infinitely long, which is another issue.)

Now, the reason that this is a toy model is that clearly we have a failure of induction. The sequence I've defined satisfies that $\varphi_0 = (0\neq 1)$ and that for each $i$, if $\varphi_i = (0 \neq 1)$, then $\varphi_{i+1} = (0\neq 1)$, yet it is not true that for all $i$, $\varphi_i = (0\neq 1)$. In fact, non-standard models of arithmetic are much harder to describe than what I've given above (and in some sense cannot be described 'explicitly'). In an actual model of $\mathsf{PA} + \neg\mathrm{Con}(\mathsf{PA})$, the proof of $0=1$ is ultimately unsound for the same reason (ill-foundedness of the implicit proof tree, or in other words a conclusion relying on an infinite regress), but the 'gap' or 'jump' is much harder to locate in some precise sense.

As is being discussed in the comments, the non-standard proofs in non-standard models of $\mathsf{PA}$ are not trustworthy (in that they're not sound). I'll give a sort of 'toy model' illustrating the intuition about why this is the case.

In a Hilbert-style deduction calculus you have some starting set of axioms and one or more deduction rules (often just modus ponens). A proof is a sequence $$ \varphi_0,\varphi_1,\dots,\varphi_n$$ of sentences satisfying that for each $i\leq n$, either $\varphi_i$ is an axiom or there are $j,k<i$ such that $\varphi_k = (\varphi_j \to \varphi_i)$.

Given such a derivation, the argument that we should believe $\varphi_n$ (provided that we believe in the axioms and in modus ponens) is ultimately an inductive proof. We argue that we should believe $\varphi_0$ and that if we believe $\varphi_i$ for all $i<j$, then we should believe $\varphi_j$.

Non-standard models are a lot more complicated than what I'm about to describe, but this really gets to the heart of the matter. Suppose we're in a non-standard model of arithmetic and the numbers look like this: $$ 0, 1, 2, \dots \infty - 2, \infty - 1, \infty, \infty + 1, \infty +2,\dots,$$ where '$\infty$' is some fixed infinite non-standard number. Now suppose I give you a derivation that lives in this model which looks like this: $$\varphi_0, \varphi_1,\varphi_{-2},\dots, \varphi_{\infty - 2},\varphi_{\infty-1}, \varphi_{\infty}.$$ This may very well be a derivation in the above sense (as in each step is either an axiom or follows from two previous steps by modus ponens), but the existence of such an object doesn't imply that $\varphi_\infty$ is a trustworthy statement. In fact (and this is the really toy part as I'll explain in a second), we could just have

• $\varphi_i = ((0= 1)\to(0 = 1))$ for each (externally) finite $i$,
• $\varphi_{\infty+i} = (0 = 1)$ for each integer $i$.

This sequence of sentences satisfies the definition (each $\varphi_i$ for $i$ finite is an axiom and $\varphi_{\infty+i}$ follows from $\varphi_0$ and $\varphi_{\infty + i -1}$ by modus ponens) but clearly the conclusion is not supported because the proof of it is relying on an infinite regress. (And this is even ignoring the possibility of sentences that are infinitely long, which is another issue.)

Now, the reason that this is a toy model is that clearly we have a failure of induction. The sequence I've defined satisfies that $\varphi_0 = (0\neq 1)$ and that for each $i$, if $\varphi_i = (0 \neq 1)$, then $\varphi_{i+1} = (0\neq 1)$, yet it is not true that for all $i$, $\varphi_i = (0\neq 1)$. In fact, non-standard models of arithmetic are much harder to describe than what I've given above (and in some sense cannot be described 'explicitly'). In an actual model of $\mathsf{PA} + \neg\mathrm{Con}(\mathsf{PA})$, the proof of $0=1$ is ultimately unsound for the same reason (ill-foundedness of the implicit proof tree, or in other words a conclusion relying on an infinite regress), but the 'gap' or 'jump' is much harder to locate in some precise sense.

As is being discussed in the comments, the non-standard proofs in non-standard models of $\mathsf{PA}$ are not trustworthy (in that they're not sound). I'll give a sort of 'toy model' illustrating the intuition about why this is the case.

In a Hilbert-style deduction calculus you have some starting set of axioms and one or more deduction rules (often just modus ponens). A proof is a sequence $$ \varphi_0,\varphi_1,\dots,\varphi_n$$ of sentences satisfying that for each $i\leq n$, either $\varphi_i$ is an axiom or there are $j,k<i$ such that $\varphi_k = (\varphi_j \to \varphi_j)$.

Given such a derivation, the argument that we should believe $\varphi_n$ (provided that we believe in the axioms and in modus ponens) is ultimately an inductive proof. We argue that we should believe $\varphi_0$ and that if we believe $\varphi_i$ for all $i<j$, then we should believe $\varphi_j$.

Non-standard models are a lot more complicated than what I'm about to describe, but this really gets to the heart of the matter. Suppose we're in a non-standard model of arithmetic and the numbers look like this: $$ 0, 1, 2, \dots \infty - 2, \infty - 1, \infty, \infty + 1, \infty +2,\dots,$$ where '$\infty$' is some fixed infinite non-standard number. Now suppose I give you a derivation that lives in this model which looks like this: $$\varphi_0, \varphi_1,\varphi_{-2},\dots, \varphi_{\infty - 2},\varphi_{\infty-1}, \varphi_{\infty}.$$ Now this may very well be a derivation in the above sense (as in each step is either an axiom or follows from two previous steps by modus ponens), but the existence of such an object doesn't imply that $\varphi_\infty$ is a trustworthy statement. In fact (and this is the really toy part as I'll explain in a second), we could just have

• $\varphi_i = ((0\neq 1)\to(0 \neq 1))$ for each (externally) finite $i$,
• $\varphi_{\infty+2i} = (0 = 1)$ and $\varphi_{\infty+2i+1} = ((0=1) \to (0=1))$ for each integer $i$.

This sequence of sentences satisfies the definition but clearly the conclusion is not supported because the proof of it is relying on an infinite regress. (And this is even ignoring the possibility of sentences that are infinitely long, which is another issue.)

Now, the reason that this is a toy model is that clearly we have a failure of induction. The sequence I've defined satisfies that $\varphi_0 = (0\neq 1)$ and that for each $i$, if $\varphi_i = (0 \neq 1)$, then $\varphi_{i+1} = (0\neq 1)$, yet it is not true that for all $i$, $\varphi_i = (0\neq 1)$. In fact, really non-standard models of arithmetic are much harder to describe than what I've given above (and in some sense cannot be described 'explicitly'). In an actual model of $\mathsf{PA} + \neg\mathrm{Con}(\mathsf{PA})$, the proof of $0=1$ is ultimately unsound for the same reason (ill-foundedness of the implicit proof tree, or in other words a conclusion relying on an infinite regress), but the 'gap' or 'jump' is much harder to locate in some precise sense.

As is being discussed in the comments, the non-standard proofs in non-standard models of $\mathsf{PA}$ are not trustworthy (in that they're not sound). I'll give a sort of 'toy model' illustrating the intuition about why this is the case.

In a Hilbert-style deduction calculus you have some starting set of axioms and one or more deduction rules (often just modus ponens). A proof is a sequence $$ \varphi_0,\varphi_1,\dots,\varphi_n$$ of sentences satisfying that for each $i\leq n$, either $\varphi_i$ is an axiom or there are $j,k<i$ such that $\varphi_k = (\varphi_j \to \varphi_j)$.

Given such a derivation, the argument that we should believe $\varphi_n$ (provided that we believe in the axioms and in modus ponens) is ultimately an inductive proof. We argue that we should believe $\varphi_0$ and that if we believe $\varphi_i$ for all $i<j$, then we should believe $\varphi_j$.

Non-standard models are a lot more complicated than what I'm about to describe, but this really gets to the heart of the matter. Suppose we're in a non-standard model of arithmetic and the numbers look like this: $$ 0, 1, 2, \dots \infty - 2, \infty - 1, \infty, \infty + 1, \infty +2,\dots,$$ where '$\infty$' is some fixed infinite non-standard number. Now suppose I give you a derivation that lives in this model which looks like this: $$\varphi_0, \varphi_1,\varphi_{-2},\dots, \varphi_{\infty - 2},\varphi_{\infty-1}, \varphi_{\infty}.$$ This may very well be a derivation in the above sense (as in each step is either an axiom or follows from two previous steps by modus ponens), but the existence of such an object doesn't imply that $\varphi_\infty$ is a trustworthy statement. In fact (and this is the really toy part as I'll explain in a second), we could just have

• $\varphi_i = ((0= 1)\to(0 = 1))$ for each (externally) finite $i$,
• $\varphi_{\infty+i} = (0 = 1)$ for each integer $i$.

This sequence of sentences satisfies the definition (each $\varphi_i$ for $i$ finite is an axiom and $\varphi_{\infty+i}$ follows from $\varphi_0$ and $\varphi_{\infty + i -1}$ by modus ponens) but clearly the conclusion is not supported because the proof of it is relying on an infinite regress. (And this is even ignoring the possibility of sentences that are infinitely long, which is another issue.)

Now, the reason that this is a toy model is that clearly we have a failure of induction. The sequence I've defined satisfies that $\varphi_0 = (0\neq 1)$ and that for each $i$, if $\varphi_i = (0 \neq 1)$, then $\varphi_{i+1} = (0\neq 1)$, yet it is not true that for all $i$, $\varphi_i = (0\neq 1)$. In fact, non-standard models of arithmetic are much harder to describe than what I've given above (and in some sense cannot be described 'explicitly'). In an actual model of $\mathsf{PA} + \neg\mathrm{Con}(\mathsf{PA})$, the proof of $0=1$ is ultimately unsound for the same reason (ill-foundedness of the implicit proof tree, or in other words a conclusion relying on an infinite regress), but the 'gap' or 'jump' is much harder to locate in some precise sense.