Is there an infinite number of logic systems? To answer that question you need to determine what the lego blocks are. To my knowledge, the only lego blocks that exist that make up logic systems are logic rules, and most logic systems have a set number of logic rules like maybe 5 or 6, but I can't think of a logic system with an infinite number of rules or a high number of rules like 100,000. So I would be inclined to believe that there are not an infinite number of rules, but the number could be as high as 1,000,000 or more. What do you think?

  • 2
    ZF set theory has infinitely many axioms. That's because two of the axioms are actually axiom schemas that stand for infinitely many axioms, one per predicate.
    – user4894
    Apr 20 at 21:01
  • 1
    Yes, and most of them are degenerate.
    – Joshua
    Apr 21 at 19:39

4 Answers 4


Two parameters that admit of infinite variability are the order of a logic, and the scale of their conjunctions/disjunctions and quantifier sets. Higher-order logic can go to e.g. a ωth order; infinitary logic can have infinitely long conjunctions/disjunctions and infinitely many quantifiers per any possible transfinite signature (though few of the possible signatures allow for "well-behaved" logics).

There is also the matter of the number of truth values a logic has; fuzzy logic has ℶ1-many, which given the indeterminacy of that expression is tantamount to arbitrarily many (for whichever value we assign to ℶ1). But then too we could have a merely ℶ0-valued logic, and then down to any finite number; so we have infinitely many options for the number of truth values we give to a logic.

Now David Lewis was of a mind to claim that the number of possible worlds could be "exactly" ℶ2, which expression is just about as indeterminate as ℶ1 already is; but so anyway, the number of possible worlds assigned to a modal logic is also a variable with infinitely many possible values.

  • Isn’t ℶ_1=2^(ℶ_0)=2^ω?
    – PW_246
    Apr 20 at 21:40
  • @PW_246 and since we have no absolute proof of what aleph that corresponds to, but can force it to be almost any aleph at all, then the expression is not really very determinate. Apr 20 at 23:56
  • It’s literally just the power set of omega. Honestly, I’d be much quicker to say that the Alephs are more dubious concepts, since we have an explicit procedure construct transfinite ordinals.
    – PW_246
    Apr 21 at 1:41
  • @PW_246 an aleph is just the cardinality of a well-ordered set, of an ordinal. Granted, it is possible to construct a system like smooth infinitesimal analysis where the trichotomy law fails, wherefore the SIA continuum is not well-ordered (since trichotomy and well-ordering are effectively interchangeable). So it would be possible to imagine that the natural powerset is not a well-ordered set. In that event, it would not be an aleph, but it would still be an (incommensurable) infinite cardinal. Not to say that the theory of cardinals is more absolute than of ordinals, though... Apr 21 at 2:11
  • 1
    Either way, I don’t see how the alephs are more sure-footed than the ordinals, since the alephs are definable from the ordinals.
    – PW_246
    Apr 21 at 2:17

I think there has been some misinterpreting the question in the results I see so far. However the question was a little unclear on a couple of points.

  • what do we mean by "logic": I assume we mean 0th or 1st order only? In the case of 2nd order logics the result is rather famously yes (see incompleteness theorems). And do you mean it has to be consistent with the ordinary model for logic?
  • what do we mean by "system": Does it have to be complete? Does it have to be decidable like regular 0th order logic? I assume it has to be consistent (self consistent, that is). And I assume you want each axiom to be independent of the others.

Well even if you answered in the more restrictive ways to those questions the answer is still "yes". You just take some system like Fitch and replace the double negative rule (ie. ¬¬p ⊢ p) with a series of larger-number-negative rules and also replace reductio ad absurdum (ie. (p→(γ∧¬γ))⊢ p ) with something that uses a larger number of nots. For instance you can use:

(12N) ¬¬¬¬¬¬¬¬¬¬¬¬p ⊢ p

(6:4N) ¬¬¬¬¬¬p ⊢ ¬¬¬¬p

(5:1RAA) (p→(γ∧¬¬¬¬¬γ))⊢ ¬p

and the proof of the double negative rule would be:

Suppose ¬¬p. Suppose BWOC ¬¬¬¬¬¬¬p. Thus (¬¬p)∧¬¬¬¬¬(¬¬p). Thus by (5:1RAA) ¬¬¬¬¬¬¬¬p. Thus by (6:4N) ¬¬¬¬¬¬p. Suppose BWOC ¬¬¬¬¬¬¬¬¬¬¬p. Thus (¬¬¬¬¬¬p)∧¬¬¬¬¬(¬¬¬¬¬¬p). Thus by (5:1RAA) ¬¬¬¬¬¬¬¬¬¬¬¬p. Thus by (12N) p. QED.

(5:1RAA) essentially allows you to add 3 pairs of nots to any statement. But adding 3 pairs to 1 pair will never result in a multiple of 6 pairs needed to use (12N). However (6:4N) allows you to reduce 3 of those pairs to 2 pairs of nots.

And now that we have proven that the double negative property in this system you can see that you can easily prove the ordinary RAA. Thus this system is equivalent to the typical Fitch system. Of course I only added 1 rule through this silliness, but you can add more. For instance consider:

(60N), (45:15N), (30:10N), (30:12N), (59:1RAA)

Which in terms of the numbers of pairs of nots accomplish {-30,-15,-10,-6,+30} respectively, all of which are needed in order to prove the double negative rule. One can see how I constructed this system from the primes {2,3,5}. And by adding more primes you can increase the number of axioms in the system. For instance the next one could have {-210,-105,-70,-42,-30,+210}. And all such systems will be equivalent to the typical Fitch system.

  • Note that 1st order logic is undecidable Apr 22 at 15:12
  • 1
    Oops, yes, how embarrassing, clearly that should have said "0th" not "1st" since that was what I was discussing. I must have just miscounted there a bit, good catch. Apr 24 at 5:24

What you have cited is a pragmatic limit, as you have not seen logic systems with more than 8 or so precepts.

  • IF there were such a limit to precept quantity,
  • then YES there would be a limit to the number of different logics.

However, there is no such limit to precept quantity.

Therefore, there are infinite possible logics.


We only know of one logic, namely human logic. This does not preclude the possibility that other logics exist, but since we don't understand yet how our own logic works, we could not tell in any event whether it would be the same as another. The fact that two logics produce the same inferences wouldn't prove that they would be identical logics.

Until we know of any other logic than our own, a system of logic, properly so called, can only be a model of our own logic.

False models are not proper models. Remains the question of true models.

The sort of axiomatic systems as used by mathematicians cannot possibly be models of our logic, and so cannot be systems of logic. A model of human logic would have to be an algorithmic model. If we presume without demonstration that there is a potential infinity of algorithmic models for any given problem, we can infer, trivially, that there has to be a potential infinity of algorithmic models of human logic.

This is clearly not the answer to your question. Your expression "logic system" presumably refers to things like so-called "First-Order Logic", "relevance logic" etc. These things are not logic systems, for the word "logic" here doesn't refer to logic to begin with.

Still, to answer your question, there is a countable infinity of axiomatic systems, so, trivially, there is a countable infinity of axiomatic systems that mathematicians are free to call "logic system", even though they are not, and cannot be, correct models of human logic and are not therefore logic systems.

  • You’re saying that literally no axiom system is correct?
    – PW_246
    Apr 21 at 21:02
  • What do you mean by an algorithmic model? Why must a model of human logic be an algorithmic model? Apr 22 at 0:31
  • @confusedcius Kudos for your name! - 2. "What do you mean by an algorithmic model?" I mean what it says in English. If you speak English, you understand what I mean. - 3. "Why must a model of human logic be an algorithmic model?" Logic is a mental capacity supported by a cognitive process. Apr 22 at 9:46
  • @PW_246 "You’re saying that literally no axiom system is correct?" Not quite. What I say is pretty clear. No axiomatic system is a model of our logic, and axiomatic systems are therefore neither true nor false in this respect. They are just irrelevant. What is false, though, is to claim, explicitly or not, that an axiomatic system is a model of logic, a system of logic, or a logic system, or that it is logic, or part of logic. What is also false is also to claim that the horseshoe is the implication (or "an" implication). Apr 22 at 9:55
  • @Speakpigeon don’t throw words around that you don’t understand. If by algorithmic you mean that our cognitive processes follow some sort of algorithm, then that’s controversial. And by the way, most ordinary English speakers don’t have a concrete idea of what an algorithm is; I doubt you do either Apr 22 at 13:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .