Does every simple first-order theory have stable forking

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Work in a sufficiently saturated model of a complete first-order theory $T$. A formula $\varphi(x,y)$ is stable if no tuples $(a_i,b_i)_{i\ge0}$ satisfy $\varphi(a_i,b_j)$ exactly when $i<j$. The theory is simple if no formula has the tree property: there are no $k\ge2$ and parameters $b_\eta$ ($\eta\in\mathbb N^{<\mathbb N}$) for which each branch gives a consistent set $\{\varphi(x,b_{\sigma\upharpoonright n}):n\ge0\}$ while each family of immediate successors is $k$-inconsistent. A formula $\psi(x,d)$ divides over $C$ if some $C$-indiscernible sequence $(d_i)$ with $d_0=d$ makes $\{\psi(x,d_i)\}$ inconsistent; it forks over $C$ if it implies a finite disjunction of dividing formulas. A type forks if it contains a forking formula.

Statement recorded in the supplied source.

In every simple theory, if $\operatorname{tp}(a/Cb)$ forks over $C$, must it contain a formula $\varphi(x,\bar b)$ which forks over $C$ and whose parameter-free formula $\varphi(x,y)$ is stable? This is the historical stable forking conjecture. The cited 2026 paper states a counterexample in Theorem 1.3. [S2]

Short English statement

Must forking in a simple theory always be witnessed by a stable formula? The cited paper claims a negative answer to this historical question.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 7600009 (AMR-075-0009). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 James Freitag and Scott Mutchnik, A Counterexample to the Stable Forking Conjecture, arXiv:2609.00436v1 (31 August 2026), Definitions 1.1 and Conjecture 1.2, p. 2; Theorem 1.3, p. 3. https://arxiv.org/pdf/2609.00436.