# Is there any way to estimate inconsistency probabilty?

Suppose one is given some language $$\mathfrak{L}$$. Is there any way to estimate the probability that a randomly chosen set of $$n$$ formulæ (all of length $$\leq k$$) is inconsistent?

No, because there are infinitely many formulae of any length for any logical language $$\mathfrak{L}$$.

However, you can prove that the set of all propositions $$\phi$$ where $$length(\phi) \leq k$$ is inconsistent if, and only if either $$1 < k$$ and $$\bot$$ is not a formula of $$\mathfrak{L}$$, or $$0 < k$$ and $$\bot$$ is a formula of $$\mathfrak{L}$$.

I'll try to give a more exhaustive take on this question given some comments from @alex-kruckman , who was willing to be more charitable than I was regarding what would qualify as a logical language.

Essentially, you'd need to meet some criteria for there to be a population of propositions from which you could reasonably infer the probability that a set of $$n$$ items of propositions $$\phi$$ of at most length $$k$$:

• An finite number of logical primitives,
• A finite number of wff-rules,
• A language containing at least one operation or formula where contradiction is possible (i.e., $$\{A, \neg A\} \subseteq S$$, where $$S$$ is the set into which propositions of $$\mathfrak{L}$$ are placed, or where it's possible that $$\bot \in S$$),
• A clear-cut decision as to what decides the length $$k$$ (e.g., whether $$\neg A$$ or $$\overline{A}$$ is how we'd define negation, and whether they count for the same or different lengths).

The smallest language that meets these criteria and for which the probability can be calculated is $$\mathfrak{L}^\bot$$, which admits only $$\bot$$, and the probability of inconsistency is 0 when $$S = \emptyset$$ and 1 when $$S \neq \emptyset$$. However, the maximum length $$k$$ is limited likewise to 0 and 1.

A nontrivial language $$\mathfrak{L}^\neg$$, let's say limited to a propositional language with primitives $$\{\bot, A\}$$ and negations on them as the only permissible wff-rules. The powerset of all well-formed formulae $$\phi$$ where $$\forall\phi(length(\phi) \leq k)$$, for every number $$k$$, defines the population.

From there, the probability that some arbitrarily selected set from that powerset is the number of sets that are inconsistent divided by the total number of sets in the powerset. If you further limit the size of each set to $$n$$ elements in each one, you can simply remove each $$S_n$$ from the powerset when $$|S_n| \neq n$$, and then perform the calculation.

##### Example:

Assume $$k = 3$$, $$n = 2$$, and the language is $$\mathfrak{L}^\neg$$.

Generate a set of all well-formed formulae from length 1 to $$k$$, and call it $$W$$. So $$W = \{\bot, A, \neg \bot, \neg A, \neg \neg \bot, \neg \neg A\}$$.

$$\mathcal{P}(W) = \{\emptyset, ..., W\}$$.

$$\mathcal{P}(W)_{|S_n| = 2} = \{\{\bot, A\}, \{\bot, \neg \bot\}, \{\bot, \neg A\}, \{\bot, \neg \neg \bot\}, \{\bot, \neg \neg A\}, \{A, \neg \bot\}, \{A, \neg A\}, \{A, \neg \neg \bot\}, \{A, \neg \neg A\}, \{\neg \bot, \neg A\}, \{\neg \bot, \neg \neg \bot\}, \{\neg \bot, \neg \neg A\}, \{\neg A, \neg \neg \bot\}, \{\neg A, \neg \neg A\}, \{\neg \neg \bot, \neg \neg A\}\}$$

$$|\mathcal{P}(W)_{|S_n| = 2}| = 15$$

Find the inconsistent subset $$I$$. $$I = \{\{\bot, A\}, \{\bot, \neg \bot\}, \{\bot, \neg A\}, \{\bot, \neg \neg \bot\}, \{\bot, \neg \neg A\}, \{A, \neg A\}, \{\neg \bot, \neg \neg \bot\}, \{\neg A, \neg \neg \bot\}, \{\neg A, \neg \neg A\}, \{\neg \neg \bot, \neg \neg A\}\}$$.

$$|I| = 10$$.

Therefore, the probability of inconsistency is $$|I|/|\mathcal{P}(W)_{|S_n| = 2}| = 10/15 = 2/3$$.

• "there are infinitely many formulae of any length" - surely it's reasonable in this context to restrict to a finite language and a finite supply of variables (greater than the length under consideration) or to count formulas up to renaming of variables. If we do this, there are only finitely many formulas of a fixed length. And the comment about the set of all formulas seems irrelevant. Commented Aug 10 at 3:09
• OP will need to specify that context, then. Otherwise, it's just as reasonable to restrict the language to one without negation or falsum, and the answer would be no, instead of yes. I suspect it's more valuable to know that, in the main, it's not possible, except over infinitely-sized sets of propositions. Commented Aug 10 at 8:41
• I agree the OP should specify the context better. Most likely they are thinking about classical first-order logic, where the "language" refers to the function and relation symbols, not to things like whether negation is included. What do you mean by "it's not possible, except over infinitely-sized sets of propositions"? Commented Aug 10 at 13:32
• Just that, if we're taking infinitely many distinct propositions of any length, the set is guaranteed to contain at least two propositions that contradict each other, or falsum. That won't satisfy the random selection part of OP's request, since it'll select all of them, but the proofs that they're inconsistent are plain. Commented Aug 10 at 19:45