02 β€” The Referee's Three Formal Lemmas

To satisfy adversarial mathematical peer review, CLUE/1.0 formalizes three explicit structural foundations:

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1. Lemma 3.1 (Prefix-Free Kraft–McMillan Admissibility Under Clue Updates)

Statement: For every vertex $u \in V$, decoded path prefix $p_i$, and clue operator $c \in \mathcal{C}$, the clue-updated transition code lengths $\ell_c(e \mid p_i) = -\log_2 P_c(e \mid p_i)$ over outgoing edges $E(u)$ satisfy the Kraft–McMillan inequality:

$$\sum_{e \in E(u)} 2^{-\ell_c(e \mid p_i)} = \sum_{e \in E(u)} P_c(e \mid p_i) \le 1$$

Consequently, prefix-free unique decodability is preserved across all clue updates $w \mapsto Cw$.

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2. Proposition 3.3 (Finite Simple-Path Route-Family Construction)

Statement: Let $G = (V, E)$ be finite with $|V| = n$. Define the admissible route family $\Gamma(s,t)$ as the set of simple (cycle-free) directed walks from $s$ to $t$ of length at most $n - 1$:

$$\Gamma(s,t) = \left\{ \gamma = (e_1, \dots, e_k) : s \leadsto t \text{ simple walk},\; k \le |V|-1 \right\}$$

After extracting closed-loop zero-boundary arbitrage cycles ($\Delta\Phi = 0$), residual edge weights satisfy $d_c(e) \ge 0$, so any walk containing a directed cycle has cost greater than or equal to the simple walk obtained by excising the cycle. Thus $|\Gamma(s,t)| < \infty$ and $\mathcal{D}(Cw)$ is non-empty.

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3. Definition 4.4 (Common Weight-Space Domain $\mathcal{W}$ for Commutator $\mathfrak{D} = [\mathcal{D}, C]$)

Statement: Define the common weight space $\mathcal{W} = \mathbb{R}_{>0}^{|E|}$. The clue operator acts as an endomorphism $C: \mathcal{W} \to \mathcal{W}$, the route-selection correspondence acts as $\mathcal{D}: \mathcal{W} \to 2^{\Gamma(s,t)}$, and the scalarized Minimum Description Length functional $L_{\text{MDL}}: \mathcal{W} \to \mathbb{R}_{\ge 0}$ is given by $L_{\text{MDL}}(w) = \min_{\gamma \in \Gamma(s,t)} L_w(\gamma)$. The Discovery Gain is the well-defined scalar functional:

$$\mathfrak{D}(w) = L_{\text{MDL}}(w) - L_{\text{MDL}}(Cw) = 12.0150\text{ bits}$$