I completed this truth table in an assignment for an implication and got a few of the outputs wrong. I was wondering if anyone can help explain why I got them wrong. This is part of the truth table that I got wrong:
PLEASE FORGIVE THE TYPO IN TABLE 4. It's meant to say $(\neg r \Rightarrow p) \land (r \Rightarrow q)$ AND NOT $(\neg p \Rightarrow p) \land (r \Rightarrow q)$
Looks like whoever marked this assumed you knew which order the alphabet went in. Notice that if you put $p,q,r$ in alphabetical order you get
$$ \begin{array}{c|c|c|c} p & q & r & (\neg r \to p) \wedge (r \to q) \\\hline 0 & 0 & 0 & 0 \\\hline 0 & 0 & 1 & 0 \\\hline 0 & 1 & 0 & 0 \\\hline 0 & 1 & 1 & 1 \\\hline 1 & 0 & 0 & 1 \\\hline 1 & 0 & 1 & \color{red}0 \\\hline 1 & 1 & 0 & \color{red}1 \\\hline 1 & 1 & 1 & 1 \end{array} $$
They were probably marking like 30 of these truth tables so they only went off the final values and didn't pay attention to the order of the inputs.
r, p, q then, and not the OPs order, wouldn't you? But anyway - usually the variable names are chosen so that alphabetic is also the most logical order, unless r is supposed to act as a shorthand as well as a variable.
$\endgroup$
Seems correct to me
\begin{array}{c:c}p & r & q & \neg r & (¬r \to p) & (r \to q) & (¬r \to p) \wedge (r \to q) \\ \hdashline 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ \hdashline 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ \hdashline 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ \hdashline 0 & 1 & 1 & 0 & 1 & 1 & 1 \\ \hdashline 1 & 0 & 0 & 1 & 1 & 1 & 1 \\ \hdashline 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ \hdashline 1 & 1 & 0 & 0 & 1 & 0 & 0 \\ \hdashline 1 & 1 & 1 & 0 & 1 & 1 & 1 \end{array}
Edit: Ah. Trevor has spotted what may have happened.
