Q: 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.

A: 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.

A: By definition, an inference rule is sound if every expression produced by the rule from a set S of expressions also logically follows from S. Using your truth table, you may show that the expression ¬P logically follows from the set S = {(P→Q),¬Q} by pointing out that every interpretation that satisfies the expressions in S, in this case only row 2, also satisfies the inferred expression ¬P.  Therefore, the inference by modus tollens is sound.
