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The formula shown in the Figure arises by substituting <math>p + \operatorname{d}p</math> for <math>p\!</math> and <math>q + \operatorname{d}q</math> for <math>q\!</math> in the boolean product or logical conjunction <math>pq\!</math> and writing the result as a cactus graph.  This follows from the fact that the boolean sum <math>p + \operatorname{d}p</math> is equivalent to the logical operation of exclusive disjunction, which parses to a cactus graph of the following form:

The cactus formula <math>\texttt{(p, dp)(q, dq)}</math> and its corresponding graph arise by substituting <math>p + \operatorname{d}p</math> for <math>p\!</math> and <math>q + \operatorname{d}q</math> for <math>q\!</math> in the boolean product or logical conjunction <math>pq\!</math> and writing the result in the two dialects of cactus syntax.  This follows from the fact that the boolean sum <math>p + \operatorname{d}p</math> is equivalent to the logical operation of exclusive disjunction, which parses to a cactus graph of the following form:

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Next question:  What is the difference between the value of the

Next question:  What is the difference between the value of the

proposition <math>pq\!</math> over there, at a distance of <math>\operatorname{d}p</math> and <math>\operatorname{d}q,</math> and the value of the proposition <math>pq\!</math> where you are satnding, all expressed in the form of a general formula, of course?  Here is the appropriate formulation:

proposition <math>pq\!</math> over there, at a distance of <math>\operatorname{d}p</math> and <math>\operatorname{d}q,</math> and the value of the proposition <math>pq\!</math> where you are standing, all expressed in the form of a general formula, of course?  Here is the appropriate formulation:

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