Can the "proves consistent" relation be proven antisymmetric?

Question

Can the "proves consistent" relation be proven antisymmetric? By antisymmetric I mean there is not some chain of theories in which some theory proves other theories consistent which in turn prove it consistent.

If $\succ$ is the "proves consistent" relation:

$A\succ B\iff (A\models Con(B))\quad\forall A\forall B $

then I think antisymmetry is usually written as follows:

$A\succ B\land B\succ A\implies A=B\quad\forall A\forall B$

Can this be proven?

This is an inquiry motivated in part by learning that Gentzen's Consistency Proof is neither weaker nor stronger than PA, which suggests an ordering may not necessarily be implied by proving consistency.

It would seem to me that by Godel's imcompleteness theorem we're denied reflexivity:

$A\succ B\implies A\neq B$

which denies us the usual antisymmetry rule. But it seems this doesn't necessarily rule out antisymmetry provided we have transitivity. Because if $\succ$ were both transitive and symmetric, we might have the following contradiction:

$A\succ C\succ A \implies A\succ A\implies A\neq A$

$\therefore\nexists C\mid A\succ C\succ A$

Which fulfils the purpose of antisymmetry when we're denied reflexivity.

So the limit of what I can prove is that "proves consistent" is antisymmetric provided it is also transitive - which I have, as yet, no reason to believe it is.

But if from $A$ proves $\text{Cons}(B)$ and $B$ proves $\text{Cons}(A)$ it follows that $A=B$, it follows also: $A$ proves $\text{Cons}(A)$. — Mauro ALLEGRANZA, Mar 29 '17 at 10:17
@MauroALLEGRANZA sorry I misread, I missed the word "from". Yes, I think what you say is stated in the question - namely that reflexivity is a contradiction and therefore the conventional definition of antisymmetry is denied us. — it's a hire car baby, Mar 29 '17 at 11:42
Your notation is wrong. If $A$ is a system (or first-order theory), it is meaningless to write "$A \implies \cdots$". Please study either Stephen's or Rautenberg's reference mentioned here. — user21820, Mar 29 '17 at 11:52
@user21820 Ok will do thanks. I had a feeling I needed "semantically encapsulates" or something like that but didn't want to go off using language I don't really have a command of. — it's a hire car baby, Mar 29 '17 at 12:02
@user21820 will replacing $\implies$ with $\models$ within the brackets resolve the notation issue (obviously it won't resolve the issue of my incompetence at logic)? — it's a hire car baby, Mar 29 '17 at 13:25
@user21820 ok well i changed it to that and hopefully that's right! — it's a hire car baby, Mar 29 '17 at 13:33
Under reasonable assumptions on the theories in question, this cannot happen - the case for a pair of theories is addressed at this question, and generalizing the answer to handle arbitrarily long "consistency cycles" is not hard. On the other hand, we do need some assumptions on the theories in question, since there are pathologically weak theories which do prove their own consistency! — Noah Schweber, Mar 29 '17 at 13:55
@NoahSchweber thanks. I've read the "informal version" for now. Does that make "proves consistent" transitive too, or does it dodge that? — it's a hire car baby, Mar 29 '17 at 14:02
@RobertFrost Assuming again that the theories are reasonable, yes, "proves consistent" is transitive. Roughly we argue as follows. Suppose $A$ proves $B$ is consistent and $B$ proves $C$ is consistent. Now by non-silliness, each theory knows all true $\Sigma^0_1$ sentences; moreover, each theory knows that the others are also non-silly (this was the content of this followup question of mine). Finally, "$X$ proves $p$" is $\Sigma^0_1$. So what? (Continued) — Noah Schweber, Mar 29 '17 at 14:53
Well, we can now argue in $A$ as follows: $B$, being nonsilly, would know that $C$ was inconsistent if $C$ were inconsistent; since $B$ proves that $C$ is consistent, that means that if $C$ is inconsistent, then $B$ is inconsistent. But $B$ is consistent (remember we're arguing inside $A$, and $A$ knows this about $B$)! So $C$ must be consistent. Getting all the details straight requires a careful checking, but this is how that goes. — Noah Schweber, Mar 29 '17 at 14:54
@NoahSchweber Thanks for use of "non silly". You know your audience! — it's a hire car baby, Mar 29 '17 at 16:18
Your notation is still wrong. "$\vdash$" and "$\vDash$" are completely different. And studying those references would make you competent in logic, which I'm going to yet again ask you to do. However, your question about transitivity of "proves consistent" is an interesting one. I've given a complete proof using provability logic, and have attempted to give dumbed-down side-explanations of each step, but you're still going to have to understand the notion of "proof verifier" from my linked post to really grasp why the steps are valid. — user21820, Mar 29 '17 at 16:36
@user21820 Thanks. I'm taking baby steps. Rome wasn't built in a day. — it's a hire car baby, Mar 29 '17 at 16:38
You're welcome! Given that your questions already have interesting core inquiries despite your current unfamiliarity with the subject, I'm sure that you would be able to go very far if you actually spent a few months to get a solid foundation. It took me some time myself plus the linked blog post by Aaronson, to reach my current understanding of these incompleteness theorems. I'm not sure how but it seems I could post an answer despite the duplicate closure; writing it took some time so I hope you can learn from it! =) — user21820, Mar 29 '17 at 16:43
@user21820 I do listen. I have to balance doing what I'm interested in with the graft of formal learning to keep myself motivated. I have a large body of material to study on p-adics, metric spaces, abstract algebra, continued fractions and Laurent series to study in support of my (probably futile) assault on the Collatz conjecture, which will come before the foundations of logic sadly so you will no doubt be n̶a̶g̶g̶i̶n̶g̶ encouraging me to do so for some time to come ;) — it's a hire car baby, Mar 29 '17 at 16:53

score 2 · Accepted Answer · edited Apr 13 '17 at 12:19

The question is ill-posed until you specify the kind of systems you have in mind, and your definition of "Con", without which you cannot ask whether a system proves "Con" of some other system... $ \def\box{\square} \def\con{\text{Con}} \def\pa{\text{PA}} $

Henceforth I shall restrict to systems with proof verifiers that interpret arithmetic (described here) including not proving the translation of "$\bot$". This is because for systems that interpret arithmetic it makes sense to ask whether they prove $\con(T)$ for any given system $T$ with a proof verifier. I shall use the standard notation from provability logic for ease of reasoning and explanation. I shall also omit the translation for the interpretation of arithmetic, but you should be able to easily fill that in yourself.

As Mauro said in a comment, it cannot be that $A \succ B \succ A$ and $A = B$ for any such systems $A,B$, due to irreflexivity of $\prec$, which is a trivial consequence of the incompleteness theorem. So clearly antisymmetry is out of the question, unless as you point out we also have transitivity of $\prec$, which would make antisymmetry vacuously true.

Theorem: $\prec$ is in fact transitive.

Take any such systems $A,B,C$.

If $A \vdash \con(B) \equiv \neg \box_B \bot$ and $B \vdash \con(C) \equiv \neg \box_C \bot$:

Within $A$:

$\neg \box_B \bot$.

$\box_B \neg \box_C \bot$. [$A$ can check that $B$ proves $\neg \box_C \bot$.]

If $\box_C \bot$: [Namely, if $C$'s verifier terminates and accepts some proof of "$\bot$".]

$\box_\pa \box_C \bot$. [$A$ can prove by induction that $\pa$ can prove that termination.]

$\box_B \box_C \bot$. [$A$ can translate the termination proof in $\pa$ to a proof in $B$.]

$\box_B \bot$. [$A$ can combine the proofs of $\box_C \bot$ and $\neg \box_C \bot$ to get a proof of $\bot$ in $B$.]

Contradiction.

Therefore $\neg \box_C \bot$.

Therefore $A \vdash \con(C)$.

Another consequence of transitivity is that there is no cycle of such theories that prove each other's consistency in a cycle.

Can the "proves consistent" relation be proven antisymmetric?

1 Answers1