(related to Lemma: Modus Tollens)
We want to prove that modus tollens is a valid logical argument.
* Modus tollenscan be formulated in propositional logic as $((p\Rightarrow q)\wedge \neg q)\Rightarrow \neg p.$
* Reformulating the implication as disjunction gives us the expression $((\neg p\vee q)\wedge \neg q)\Rightarrow \neg p.$
* On the left side we have only the operations "$\wedge$" and "$\vee$". Therefore, we can use the fact that propositional logic is a Boolean algebra $(B,\wedge,\vee,1,0)$, and make use of the properties of this Boolean algebra $B$ as follows:
$$\begin{array}{rl}
((\neg p\vee q)\wedge \neg q)\Rightarrow \neg p&\\
((\neg p \wedge\neg q)\vee (q\wedge \neg q))\Rightarrow \neg p&(\text{distributivity of }"\wedge"\text{ and }"\vee")\\
((\neg p \wedge\neg q)\vee 0)\Rightarrow \neg p&(\neg q\text{ is the complement element of }q)\\
(\neg p \wedge\neg q)\Rightarrow \neg p&(0\text{ is the smallest element of the$B$})\\
\end{array}$$
* It follows from the truth table of conjunction that if $\neg p$ and $\neg q$ are true, than $\neg p$ is true.
* By definition of a valid logical argument, modus tollens $((p\Rightarrow q)\wedge \neg q)\Rightarrow \neg p$ is valid.
