by $n_{qz}$, the exact test for the speciousness reduces to $$p_B=\frac{\sum_{j\geq n_{q\neg zc}}\left(n_{q\neg z} \atop j
  \right)\left(n_{\neg q} \atop n_{\neg qc}-j\right)}{\left(n-n_{qz} \atop n_c-n_{qzc}\right)}.$$

With mutual information, the equivalent test is $MI_S(Q\rightarrow
C|QZ\rightarrow C)=MI(Q\rightarrow C|\neg(QZ))=MI(\neg Q\rightarrow
\neg C|\neg(QZ)) $ (assuming $\delta_2>0$). This is the same as the significance of productivity of $\neg Q\rightarrow \neg C$ with respect to $\neg (QZ)\rightarrow \neg C$.

A further connection to redundancy testing occurs if we compare a rule $X\rightarrow C$ and its specialization $XZ\rightarrow C$ using speciousness tests. If $MI_S(X\rightarrow C|XZ\rightarrow C)=MI(\neg
X\rightarrow \neg C|\neg (XZ))$ is too small, then $X\rightarrow C$ is specious, and if $MI_S(XZ\rightarrow C|X\rightarrow
C)=MI(XZ\rightarrow C|X)$ is too small, then $XZ\rightarrow C$ is specious. Since both cannot occur simultaneously, only the rule with a smaller conditional $MI$ value can be specious. It turns out that this is the same as comparing the unconditional $MI$ values of rules. Thus, the test for redundancy in [@kingfkais] (i.e., testing if $MI(XZ\rightarrow C)<MI(X\rightarrow C)$) is a special case of speciousness testing. It is also equivalent to comparing the significance of productivity of $XZ\rightarrow C$ with respect to $X\rightarrow C$ and of $\neg X\rightarrow \neg C$ with respect to $\neg(XZ)\rightarrow \neg C$.

Equivalent rules
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A 