I just started logic and am having trouble working through it.
Can somebody show, step by step, how the following formulas are equivalent? I.e.
$(\neg q\rightarrow\neg p)\rightarrow (\neg q\rightarrow p)\rightarrow r\equiv (q \rightarrow r)$
Thank you so much!
I assume that you are working in classical propositional logic. Two formulas $\phi$ and $\psi$ are equivalent(semantically) iff they have the same truth-values under every interpretation of the variables.
I also assume that by $(\neg q\rightarrow\neg p)\rightarrow (\neg q\rightarrow p)\rightarrow r$ you mean $(\neg q\rightarrow\neg p)\rightarrow ((\neg q\rightarrow p)\rightarrow r)$
We might check this condition by building a so called truth table to list all the possible variants of such a mapping:
\begin{array}{|c|c|c|c|} \hline p& q& r & q\rightarrow r & \neg q\rightarrow\neg p & \neg q\rightarrow p & (\neg q\rightarrow p)\rightarrow r & (\neg q\rightarrow\neg p)\rightarrow ((\neg q\rightarrow p)\rightarrow r)\\ \hline 0& 0& 0& 1& 1& 0& 1& 1\\ \hline 0& 0& 1& 1& 1& 0& 1& 1\\ \hline 0& 1& 0& 0& 1& 1& 0& 0\\ \hline 0& 1& 1& 1& 1& 1& 1& 1\\ \hline 1& 0& 0& 1& 0& 1& 0& 1\\ \hline 1& 0& 1& 1& 0& 1& 1& 1\\ \hline 1& 1& 0& 0& 1& 1& 0& 0\\ \hline 1& 1& 1& 1& 1& 1& 1& 1\\ \hline \end{array}
Note that I've added some extra columns to make the calculations more clear. You only need to compare the 4-th and the last column for equivalence.
It is a good training in truth tables to construct this one yourself. Maybe check in the truth tables of the standard operators $\rightarrow$ and $\neg$ and try it yourself.
EDIT: A little note on why it suffices to check that they have the same truth value under every interpretation.
Two formulas $\phi$ and $\psi$ are equivalent if $\models\phi\leftrightarrow\psi$, that is if $\phi\leftrightarrow\psi=\phi\rightarrow\psi\land\psi\rightarrow\phi$ is a tautology, i.e. it is always true.
Now, you can check by the semantic evaluation of $\rightarrow$ and $\land$(the corresponding truth tables), that given an assignment of truth values(an interpretation) $\mathcal{I}$, that $\mathcal{I}(\phi\leftrightarrow\psi)$=1 iff $\mathcal{I}(\phi)=\mathcal{I}(\psi)$.