Conditional Probability Equivalence I am reading this paper here: http://www.utm.utoronto.ca/~weisber3/articles/SobervBJPS3.SP.pdf which claims on page 10 that $p(E \wedge [H_x \wedge \sim R]) + p(E \wedge [H_x \wedge E])= p(H_x \wedge E)$. I am having trouble seeing this equivalence.

Any help is appreciated; thank you in advance.
It doesn't.  That step is erroneous, as is the one prior (the sum-rule is only valid when the two events are mutually exclusive, which they clearly are not).  

However, both are also unnecessary, as: $[E\wedge [H_x\wedge \neg R]] \vee [E\wedge [H_x\wedge E]]) = E\wedge H_x$ , thus you immediately have the desired result.

$$p([E\wedge [H_x\wedge \neg R]] \vee [E\wedge [H_x\wedge E]])
\\[2ex] = p([[E\wedge H_x]\wedge \neg R]\vee [[E\wedge H_x]\wedge E]) 
\\[2ex] = p([[E\wedge H_x]\wedge \neg R]\vee [E\wedge H_x]) 
\\[2ex] = p([E\wedge H_x]) 
$$
It doesn't.  That step is erroneous, as is the one prior (the sum-rule is only valid when the two events are mutually exclusive, which they clearly are not).  

However, both are also unnecessary, as: $[E\wedge [H_x\wedge \neg R]] \vee [E\wedge [H_x\wedge E]]) = E\wedge H_x$ , thus you immediately have the desired result.

$$p([E\wedge [H_x\wedge \neg R]] \vee [E\wedge [H_x\wedge E]])
\\[2ex] = p([[E\wedge H_x]\wedge \neg R]\vee [[E\wedge H_x]\wedge E]) 
\\[2ex] = p([[E\wedge H_x]\wedge \neg R]\vee [E\wedge H_x]) 
\\[2ex] = p([E\wedge H_x]) 
$$