Gödel's original proof of the Completeness Theorem for first-order logic proves it in the equivalent form :

a formula $\varphi$ is satisfiable or $\varphi$ is refutable (i.e.$\vdash \lnot \varphi$).

Assuming classical logic, someone can exhibit an example of first-order formula $\varphi$ for which we are not able to decide if it is provable (i.e.$\vdash \varphi$) or if there is a counterexample (i.e.$\lnot \varphi$ is satisfiable) ? In other words, a formula $\varphi$ for which we are not able to decide if it is valid or not ?

Added March 1st

Following Carl's answer, an example of a sentence like the above can be costructed in the following way; ref to Elliott Mendelson, Introduction to Mathematical Logic (4th ed - 1997), Ch.4: Axiomatic Set Theory [page 225-on] :

(i) let $\mathcal N$ the formula obtained by the conjunction of the following axioms for $\mathsf {NBG}$ : Axiom T, Axiom D, Axiom N, Axioms B1-B7, Axiom U, Axiom P, Axiom S, Axiom I, Axiom AC, Axiom Reg (no axiom shemas), after having "unwinded" all the set-theoretic definitions (like : $\emptyset$ and $\subseteq$) and after having replaced in the above formulas all occurences of $\in$ with a binary predicate symbol $R$;

(ii) let $RH$ the formula expressing in the language of $\mathsf {NBG}$ the Riemann Hypothesis, subject to the same "translation".

Then :

$\mathcal N \land \lnot RH$

is an example of formula for which we do not know if it is satisfiable or if it is refutable.

  • $\begingroup$ The Completeness Theorem states that every sentence in first order logic is decidable. So, in principle, there is no sentence of the form you describe. However, as a practical problem, Satisfiability (assignment of truth values to variables making an arbitrary conjunction of disjunctions true) is notoriously difficult. $\endgroup$ – Eric Towers Feb 28 '14 at 8:37
  • $\begingroup$ Are you sure ? ... see Wiki Decidability of a logical system : "First-order logic is not decidable in general; in particular, the set of logical validities in any signature that includes equality and at least one other predicate with two or more arguments is not decidable". $\endgroup$ – Mauro ALLEGRANZA Feb 28 '14 at 8:39
  • $\begingroup$ You are correct. I misremembered the Completeness Theorem. It applies to first order predicate calculus, not to first order logic(s) in general. $\endgroup$ – Eric Towers Feb 28 '14 at 8:47
  • $\begingroup$ Wouldn't every (first order) independent statement qualify? For instance, do you allow the continuum hypothesis as an example? (It's neither provable nor disprovable in ZFC.) $\endgroup$ – Eric Towers Feb 28 '14 at 8:54
  • $\begingroup$ About $ZFC$, I'm not sure to grasp your argument ... assuming that I'm able to form the "conjunction" of $ZFC$'s axioms, call it $Z$, from the fact that $Z \nvdash CH$, I will derive $\nvdash Z \rightarrow CH$; your suggestion is to conclude that $\vdash \lnot (Z \rightarrow CH)$ ? $\endgroup$ – Mauro ALLEGRANZA Feb 28 '14 at 9:02

Decidability - which means computability in this question - is always about a set of formulas. It does not make much sense to look for a single formula the satisfiability of which is not decidable.

This is because, if the formula is satisfiable, then the program that always outputs "Yes" will be correct, and if the formula is not satisfiable, the program that always output "No" will be correct. So, in a somewhat trivial way, the satisfiability of each individual formula is decidable. But this is so trivial that we don't view it as interesting.

What is true is that, when we consider sufficiently rich languages of first-order logic, there is no single program that, given an arbitrary formula, will decide whether that formula is satisfiable or not. The key point is that this is a problem involving an infinite set of formulas; the same program would have to decide correctly whether each of them is satisfiable.

It is still the case, of course, that there are individual statements of first-order logic such that we do not know whether those statements are satisfiable or not. For example, let $C$ be any unsolved conjecture, expressed in the language of set theory. Let $A$ be the conjunction of the finite set of axioms for NBG set theory (which is stronger than ZFC set theory). Then we will not know whether the formula $A \land \lnot C$ is satisfiable or not. We could take $C$ to be the Riemann hypothesis or any other unsolved problem that can be formalized in set theory.

The key point is that there are foundationally strong theories that have finite axiom systems, so we can code the entire axiom system as a single formula. That is also the key fact behind the proof that first-order logic is not decidable when the language is sufficiently rich.

  • $\begingroup$ I do not understand - perhaps the "decidability" in the title of the question is wrong- The point is about my reading of G's Theorem. (i) is my reading wrong ? (I think not). (ii) if not, I've never met in a textbook the example of complicated formula with the comment : "up to now nobody knows if this formula is valid or not" ... This is my "silly" question. $\endgroup$ – Mauro ALLEGRANZA Feb 28 '14 at 11:33
  • $\begingroup$ So what you want is just an example of a formula such that nobody knows whether it is valid or not? That I can give. But in the formalism of computability theory, we know there is nevertheless a computable function that, given that formula, correctly tells whether the formula is provable or not - we just don't know which function it is. When people write about "decidability of logic" they mean a computable function that tells whether arbitrary formulas are provable. $\endgroup$ – Carl Mummert Feb 28 '14 at 11:40
  • $\begingroup$ Ok, now I've understood; my question was not about a (impossible) solution to the decidability problem... You consider $NBG$ taht is finitely axiomatizable (where $ZFC$ is not) and build the sentence $NBG \rightarrow "Conjecture"$ ... Regarding this example, I think that the "solution" proposed in a comment above (i.e.to use CH) does not work (replacing of course $ZFC$ with $NBG$) : is it right ? $\endgroup$ – Mauro ALLEGRANZA Feb 28 '14 at 12:27
  • $\begingroup$ We know that NBG + CH is satisfiable but is not valid. So it depends on whether you want an example that is satisfiable, or an example that is valid. $\endgroup$ – Carl Mummert Feb 28 '14 at 12:55
  • 1
    $\begingroup$ Ok: so I'm right. The example proposed above (with $NBG$ in place of $ZFC$, i.e.$NBG \rightarrow CH$) does not work, as I imagined. We know it is satisfiable : so $\lnot (NBG \rightarrow CH)$ is not provable ... $\endgroup$ – Mauro ALLEGRANZA Feb 28 '14 at 13:06

Just replace $\varphi$ with $\lnot \varphi$ in the quoted statement. So:

a formula $\lnot \varphi$ is satisfiable or $\lnot \varphi$ is refutable ($\vdash \lnot \lnot \varphi$).

The point is that $\lnot \lnot \varphi$ and $\varphi$ are logically equivalent in classical logic.

That said, it is difficult to decide which of the two cases happens for a particular formula $\varphi$. Indeed, this is equivalent to being able to decide whether or not a given first-order theory is consistent.

  • $\begingroup$ I'm not sure of grasping your comment. I'll try: consider a "theory" $\mathcal T = \{ \varphi \}$; now, $\varphi$ is sat iff $\mathcal T$ has a model, and $\varphi$ is refut (i.e.$\vdash \lnot \varphi$) iff $\mathcal T$ is incons. So, Godel's alternative bolis down to : $\mathcal T$ has a model or $\mathcal T$ is inconsistent... I'm right ? $\endgroup$ – Mauro ALLEGRANZA Feb 28 '14 at 11:15
  • $\begingroup$ Yes, basically. $\endgroup$ – Zhen Lin Feb 28 '14 at 11:39
  • $\begingroup$ @Carl Mummert - So, in order to exhibit the $\varphi$, I can consider a finite axiomatizable theory (following Carl's answer: $NBG$), and write the conjunction of the axioms: $N$. So, I do not know which of the alternatives holds: is $N$ satisfiable or we have $\vdash N \rightarrow \lnot N$ ? $\endgroup$ – Mauro ALLEGRANZA Feb 28 '14 at 12:48

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