is that some special cases can be checked with less computation. For example, there is no need to evaluate frequency $n_{xq}$ if the rules are equivalent or $X\subset Q$, assuming that $n_x$, $n_{xc}$, $n_q$, and $n_{qc}$ are available. Birch’s exact test is also computationally demanding and is preferably performed only when needed.

The first case covers equivalent rules which cannot be considered as specious per se. Still, they contain redundant information which only complicates understanding. If $X$ and $Q$ are equivalent, then each $X$’s rule has an equivalent $Q's$ rule; if $X$ and $\neg Q$ are equivalent, then for each $X\rightarrow C{=}c$, there is $Q\rightarrow
C{\neq}c$. Therefore, it is sufficient to report the equivalence information ($X\rightarrow Q$ or $X\rightarrow \neg Q$), unless it occurs as a dependency rule in the list (which happens when $|X|=1$ or $|Q|=1$).

The second case covers rules of the form $Q\subset X$, where $X\rightarrow C{=}c$ is a non-redundant specialization of $Q\rightarrow C{=}c$. The opposite case where $X\subset Q$ has already been handled during the search. Since $\delta_1=0$, one could simply check whether $\delta_2\leq 0$. However, evaluating significance of speciousness detects also cases where $\delta_2$ is weakly positive.

The third case corresponds to Yule-Simpson’s paradox, where 