Counterexample structure (l) — In a product of two magmas with at least two elements each, no point where E255 fails has the congruence property 3 of (k)
ProvedFiniteMagmaE677.not_pair_critical_of_prodLet and be magmas, each with at least two elements, and let be the componentwise operation on :
Let with , where . A congruence of is a reflexive, symmetric and transitive relation such that and give . Then it is not true that every congruence of that relates two distinct elements relates and . Neither magma is assumed to be finite or to satisfy E677.
So property 3 of Counterexample structure (k) fails at each point of such a product at which E255 fails. For a counterexample search: a product of two magmas with at least two elements each has no point with properties 1 and 3 of (k). This applies to the product of item 1 of the mission description; there has at least two elements, because . S. Chuks-Onah (ETP Zulip thread "FINITE: 677 -> 255", 2026-06-10) lists, among the known facts about a minimal counterexample : "E255 fails componentwise, so is not a direct product". The theorem is the case of two factors of the standard fact that a subdirectly irreducible algebra is directly indecomposable (S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, Theorem II.8.5), and the proof is the standard one with the kernels of the two projections. It is not a new result.
Formalization Note The product operation is an operation op on α × β with op p r = (op₁ p.1 r.1, op₂ p.2 r.2) for all p and r; Nontrivial α says that α has two distinct elements. The conclusion is the negation of property 3 of (k) at z, written in the same form. The theorem is about the product type itself; a magma that is only isomorphic to a product needs a transport step that is not part of this statement.
import Definitions.Def_FiniteMagmaE677 import Mathlib universe u v open FiniteMagmaE677
theorem FiniteMagmaE677.not_pair_critical_of_prod
{α : Type u} {β : Type v} [Nontrivial α] [Nontrivial β]
(op₁ : α → α → α) (op₂ : β → β → β) (op : α × β → α × β → α × β)
(hop : ∀ p r : α × β, op p r = (op₁ p.1 r.1, op₂ p.2 r.2))
(z : α × β) (hz : op (op (op z z) z) z ≠ z) :
¬ ∀ r : α × β → α × β → Prop, Equivalence r →
(∀ a a' b b' : α × β, r a a' → r b b' → r (op a b) (op a' b')) →
∀ a b : α × β, a ≠ b → r a b → r z (op (op (op z z) z) z) := by sorry