Is there any way to estimate inconsistency probabilty?

Question

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?

Answer 1

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 $\mathfrak{L}$.

Addendum:

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 $|I|/|\mathcal{P}(W)_{|S_n| = 2}| = 10/15 = 2/3$.