# Showing that Modus Tollens is sound

When asked to show that Modus Tollens is sound in the propositional calculus, I tried to do this by enumerating all interpretations using a truth table. However I am unsure that my deductions are correct:

$\begin{array}{cc|ccc} P&Q&P\to Q&\overline{Q}&\overline{P}\\ \hline T&T&T&F&F\\ F&F&T&T&T\\ T&F&F&T&F\\ F&T&T&F&T \end{array}$

My understanding is the Modus Tollens is sound, because under the interpretation when $\neg Q$ (rows 1 and 4) and when the implication is true (rows1 and 4), then we can infer $\neg P$. For rows 1 and 2, P is T and F respectively, and the negation here also holds.

I feel that this is insufficient. But I am unsure as to what I am missing.

• @Senex I have further edited your table. Here is revision history and link to your revision. Since you had | in the source but it was not displayed in the post, I guess that this is what you intended to do. But please, check it if you have time. (And if I somehow unintentionally changed what you wanted to get, please, edit the post again.) – Martin Sleziak Jul 4 '16 at 7:52
• You should refer to row 2 only. – Doug Spoonwood Jul 4 '16 at 14:02

We're missing two, so using your table:

\begin{align*}&(P\rightarrow Q)\wedge\neg Q\\&\;\;\;\;\;\;\;\;\;\;F\\&\;\;\;\;\;\;\;\;\;\;T\\&\;\;\;\;\;\;\;\;\;\;F\\&\;\;\;\;\;\;\;\;\;\;T\end{align*}

and finally

\begin{align*}&\left[(P\rightarrow Q)\wedge\neg Q\right]\rightarrow\neg P\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;T\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;T\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;T\\&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;T\end{align*}

and we get tautology.

• No, modus tollens doesn't involve a conjunction. It involves two propositions. – Doug Spoonwood Jul 4 '16 at 14:00
• @DougSpoonwood As you can check wherever you like, it is the same: modus tollens requires that $\;P\rightarrow Q\;$ and $\;\neg Q\;$ are true so that $\;\neg P\;$ will be true. – user351910 Jul 4 '16 at 14:11
• @AntoineNemesioParras I think is correct, and I've seen this form of Modus Tollens in a couple of textbooks. One other question, with an inference rule, are we trying to infer the value of a proposition or it's Truth? By that I mean, do we care about inferring only Truth or whatever the proposition is, either T or F? --edit, 'sound' means true under every interpretation ... right? – user1658296 Jul 4 '16 at 15:48
• @user1658296 I've seen it in both equivalent forms, indeed. In this special case we're trying to prove that whenever $\;P\rightarrow Q\;$ and $\;\neg Q\;$ are given, then we can infere $\;\neg P\;$ . As shown above (in the second table) , no matter what the truth values of the atoms $\;P,Q\;$ are, we get a tautology. – user351910 Jul 4 '16 at 16:36
• No, it's not the same. Modus tollens requires that (P -> Q) as well as $\lnot$P hold true. That means that we have two propositions holding true. What you wrote in the post consists of a single proposition with $\land$. $\land$ might not exist in the language, and there might not exist any other symbol for conjunction. We might have a system with just $\lnot$ and -> without any definitions, or work in the definition-free part of the language. – Doug Spoonwood Jul 5 '16 at 1:50

"Combine" your tables for ${P}\rightarrow {Q}$ and $\neg{Q}$ using "and". The result must be the same to $\neg P$.

• No, modus tollens doesn't involve a conjunction. It involves two propositions. – Doug Spoonwood Jul 4 '16 at 14:00