Editing Modus tollens (section)

===Probability calculus===
''Modus tollens'' represents an instance of the [[law of total probability]] combined with [[Bayes' theorem]] expressed as:

<math display="block">\Pr(P)=\Pr(P\mid Q)\Pr(Q)+\Pr(P\mid \lnot Q)\Pr(\lnot Q)\,,</math>

where the conditionals <math>\Pr(P\mid Q)</math> and <math>\Pr(P\mid \lnot Q)</math> are obtained with (the extended form of) [[Bayes' theorem]] expressed as:

<math display="block">\Pr(P\mid Q) = \frac{\Pr(Q \mid P)\,a(P)}{\Pr(Q\mid P)\,a(P)+\Pr(Q\mid \lnot P)\,a(\lnot P)}\;\;\;</math> and  <math display="block">\Pr(P\mid \lnot Q) = \frac{\Pr(\lnot Q \mid P)\,a(P)}{\Pr(\lnot Q\mid P)\,a(P)+\Pr(\lnot Q\mid \lnot P)\,a(\lnot P)}.</math>

In the equations above <math>\Pr(Q)</math> denotes the probability of <math>Q</math>, and <math>a(P)</math> denotes the [[base rate]] (aka. [[prior probability]]) of <math>P</math>. The [[conditional probability]] <math>\Pr(Q\mid P)</math> generalizes the logical statement <math>P \to Q</math>, i.e. in addition to assigning TRUE or FALSE we can also assign any probability to the statement. Assume that <math>\Pr(Q) = 1</math> is equivalent to <math>Q</math> being TRUE, and that <math>\Pr(Q) = 0</math> is equivalent to <math>Q</math> being FALSE.  It is then easy to see that <math>\Pr(P) = 0</math> when <math>\Pr(Q\mid P) = 1</math> and <math>\Pr(Q) = 0</math>. This is because <math>\Pr(\lnot Q\mid P) = 1 - \Pr(Q\mid P) = 0</math> so that <math>\Pr(P\mid \lnot Q) = 0</math> in the last equation. Therefore, the product terms in the first equation always have a zero factor so that <math>\Pr(P) = 0</math> which is equivalent to <math>P</math> being FALSE. Hence, the [[law of total probability]] combined with [[Bayes' theorem]] represents a generalization of ''modus tollens''.<ref>Audun Jøsang 2016:p.2</ref>