Modus tollens - Negations on the implication

Question

This is likely a basic question however based on my textbook definition of Modus tollens it looks like this:

$$\neg q$$ $$\frac{(p \implies q)}{\neg p}$$

I however have something that looks like this:

$$\neg q \implies \neg p$$ $$p$$

Would this give the conclusion of $\neg q$ or just $q$? I am confused what the result would be. With the textbook way it looks like it would be negative, however in that case the implications are not negative, the single premise is negated.

Answer 1

The contrapositive of an implication is an equivalent statement to the implication.

$$(p\implies q) \quad \equiv \quad (\neg q \implies \neg p)$$

That is $q$ is true whenever $p$ is true, if and only if $p$ is false whenever $q$ is false.

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As per the comment below:

the propositions are p is: "George has eight legs" and q is: "George is a spider". I am told the implication "If George does not have eight legs, then he is not a spider" so I made it what you see in the question with the negations. I am assuming from this you can use the contraposition and say p⟹q but I could be wrong which is why I asked the question

!   Then what you have is:

$$\begin{array}{|c}\text{Not (George is not a spider)}\\ \text{George does not have eight legs} \to \text{George is not a spider} \\ \hline \text{Not (George does not have eight legs)}\end{array} \\[1ex]\Updownarrow\\[1ex] \begin{array}{|c}\neg(\neg q)\\ (\neg p) \to (\neg q) \\ \hline \neg(\neg p)\end{array}\\[1ex]\Updownarrow\\[1ex] \begin{array}{|c}q\\ \neg p \to \neg q \\ \hline p\end{array}$$