costCompose_fourfold_power_counterexample
plain-language theorem explainer
The cost composition operation induced by the Recognition Composition Law fails power-associativity already for four identical unit elements. Researchers examining the algebraic consequences of the RCL for cost values would cite this explicit counterexample to bound the monoid-like structure. The verification is a direct numerical reduction that substitutes the closed-form definition of the operation and simplifies both sides.
Claim. Let $a ★ b = 2ab + 2a + 2b$ denote cost composition on the reals. Then $(((1 ★ 1) ★ 1) ★ 1) ≠ ((1 ★ 1) ★ (1 ★ 1))$.
background
Cost composition is the binary operation on real numbers induced by the Recognition Composition Law: given costs $a = J(x)$ and $b = J(y)$, one sets $a ★ b = 2ab + 2a + 2b$. This operation is commutative by construction and appears in the CostAlgebra module after the basic properties of the J-cost function. The shifted cost $H(x) = J(x) + 1$ is introduced immediately afterward to convert the RCL into d'Alembert's functional equation $H(xy) + H(x/y) = 2 H(x) H(y)$. Upstream definitions trace the cost function to derived costs on multiplicative recognizers and to the J-cost of recognition events.
proof idea
The proof is a one-line wrapper that applies the definition of cost composition and evaluates both sides numerically at the concrete value 1.
why it matters
The declaration supplies a concrete boundary on the algebraic structure generated by the RCL inside the Recognition Science framework. It shows that full power-associativity fails at four factors while weaker properties (commutativity, flexible associativity) continue to hold, thereby delimiting the monoid-like behavior before the module shifts to the H-monoid and the eight-tick octave. The result sits in the CostAlgebra development that precedes the forcing chain steps T5–T8 and the derivation of spatial dimension D = 3.
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