$p_{xc}=p_{xqc}+p_{x\neg qc}=(p_{xq}+p_{x\neg q})p_c=p_xp_c$, which leads to the same triplet $(p_xp_c,p_qp_c,p_{xq}p_c)$.

Let $n$, $n_c$, $n_x$, $n_q$, and $n_{xq}$ be given. Then for the probability distributions of $n_{xc}$ and $n_{qc}$ given null hypotheses holds $$\begin{gathered}
P(n_{xc}|H_{01})P(n_{qc}|n_{xc},H_{03}\wedge H_{04})=\\
P(n_{xc},n_{qc}|H_{01}\wedge H_{03}\wedge H_{04})=
P(n_{xc},n_{qc}|H_{02}\wedge H_{05}\wedge H_{06})\\
=P(n_{qc}|H_{02})P(n_{xc}|n_{qc},H_{05}\wedge H_{06})\end{gathered}$$ and thus $$\label{probratio}
\frac{P(n_{xc}~|~H_{01})}{P(n_{qc}~|~H_{02})}=
\frac{P(n_{xc}~|~n_{qc},H_{05}\wedge H_{06})}{P(n_{qc}~|~n_{xc},H_{03}\wedge H_{04})}.$$

Equation \[probratio\] means that for judging whether $P(n_{xc}~|~n_{qc},H_{05}\wedge H_{06})<P(n_{qc}~|~n_{xc},H_{03}\wedge H_{04})$ (i.e., if $r_2$ could make $r_1$ specious), it suffices to check whether $P(n_{xc}~|~H_{01})<P(n_{qc}~|~H_{02})$. This can be done by fixing the sampling model and estimating the point probabilities either directly or indirectly. However, it is not necessary to estimate the exact probabilities, because most goodness measures for statistical dependencies reflect the same distributions under the same null hypotheses. We will shortly show that for speciousness detection, it suffices that the goodness measure and its conditional counterpart are [*order-homomorphic*]{}, as defined in the following:

Let $M$ be a goodness measure for dependency rules and $M_c$ its conditional variant, such that $M(Q\rightarrow C{=}c)$ measures marginal dependence between $Q$ and $C{=}c$ and $M_c(Q\rightarrow C{=}c|X\rightarrow C{=}c_x)$ measures conditional dependence between $Q$ and $C{=}c$ given $X\rightarrow C{=}c_x$. $M$ and $M_c$ are called [*order-isomorphic*]{}, if for all rules $r_1$, $r_2$ $$M_c(r_1|r_2)\leq M_c(r_2|r_1) \Leftrightarrow M(r_1)\leq M(r_2)$$ and [*order-homomorphic*]{} 