# Proof

(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: $$\begin{array}{rl} ((\neg p\vee q)\wedge \neg q)\Rightarrow \neg p&\\ ((\neg p \wedge\neg q)\vee (q\wedge \neg q))\Rightarrow \neg p&(\text{distributivity of }"\wedge"\text{ and }"\vee")\\ ((\neg p \wedge\neg q)\vee 0)\Rightarrow \neg p&(\neg q\text{ is the complement element of }q)\\ (\neg p \wedge\neg q)\Rightarrow \neg p&(0\text{ is the smallest element of theB})\\ \end{array}$$ * 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.

∎

 Thank you to the contributors under CC BY-SA 4.0!  Github: References Bibliography Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016 
 Feeds Acknowledgments Terms of Use Privacy Policy Disclaimer © 2014+ Powered by bookofproofs, All rights reserved.