Then dependency rule $r_1$ can be made specious only by such $r_2$ for which $M(r_1)<M(r_2)$ ($M(r_1)>M(r_2)$).

The proof follows directly from the definitions of speciousness and order-homomorphism.

This has an important consequence from an algorithmic point of view. For simplicity, we give the result only for increasing goodness measures.

Let $R$ be a set of binary attributes and $\mathcal{U}$ a universe of all possible rules $X\rightarrow C{=}c$, $C\in R$, $X\subseteq R\setminus\{C\}$, $c\in \{0,1\}$. Let $(M, M_c)$ be an order-homomorphic pair of increasing goodness measures and $\tau$ a minimum threshold for $M$ (alternatively, $M$-value of the $K$th best rule). Further, let $\mathcal{R_{\tau}}=\{r~|~r\in \mathcal{U}, M(r)\geq \tau\}$ denote the set of all rules having sufficiently good $M$-value in the given data. Then, for detecting speciousness of any $r_1\in \mathcal{R}_{\tau}$, it suffices to check $r_2\in \mathcal{R}_{\tau}$, $r_2\neq r_1$, having $M(r_2)\geq M(r_1)$. Furthermore, $r_1$ (or its reverse rule, with the antecedent and consequent reversed) and $r_2$ (or its reverse rule) should have the same attribute as the consequent.

In practice, this means that speciousness detection can be done in the post-processing phase, when only the best $K$ rules or all sufficiently good rules are available. The only requirement is that the results set $\mathcal{R}$ contains 