\begin{document} \maketitle \section{Introduction} In this (unfinished) lesson on material implication, as explained using texPad, we piggy-back two important issues in this course. \subsection{Implication is non-associative} It makes a difference where you put the parenthese when you have two implications in a row. Sometimes, there are as many as five. If you understand this lesson, you can work out the more complicated case. \subsection{Using texPad} For many reasons, and even more tomorrow, we must find substitutes for the traditional slate and chalk (or, whiteboard and marker) expository tool in mathemetatics. The overriding principle to apply in judging such a substitute is whether or not the technology makes its use no more cumbersome and whether it is ubiquitous. Our version of this is texPad (see Advice/ for more details.) \subsection{The argument} In the first panel we showed that if $A,B,C$ are propositions then the compound proposition $D = (A\implies (B \implies C))$ is equivalent to $E = (A \and B) \implies C$. This reformulation makes it easy to understand, use, and even prove by contradiction. One of the intermediary steps was \begin{eqnarray*} A \implies (B \implies C) & \sim & \bar{A} \or \bar{B} \or C \\ \therefore \neg (A \implies (B \implies C)) & \sim & A \and B \and \bar{C} \\ \end{eqnarray*} Thus, to prove the first one true, the very least you need to do is show that either $A$ is false, $B$ is false, or $C$ is true. In the second panel, we simplify $(A \implies B) \implies C$ as much as we can to get $(A \and \bar{B})\or C$. This does tell us that, for this double implication to be true, either $C$ must already be known to be true, or both $A$ must be true and $B$ must be false. Since we end up with a disjunction of $C$ with a conjunction $A \and \bar{B}$ we can go a little further. The $\and$ and $\or$ operators are each distributive across the other. (See the textbook or google it.) That means we can rewrite \begin{eqnarray*} (A \and \bar{B})\or C&\sim & (A\or C) \and (\bar{B} \or C) \\ & \sim & (A\or C) \and (B \implies C) \\ & \sim & (\bar{A} \implies C) \and (B \implies C) \\ \end{eqnarray*} for what that's worth. Perhaps a lawyer could use this equivalence to really complicate a legal document. \section{Conclusion} It's easy to forget what the purpose of this lesson was, namely to show that the two double implications are not the same. What makes material implication so special is that there is only one 0 in it's truth table. That is, only $(1 \implies 0)$ is false. All 3 other cases are true. This suggests a quick test \begin{eqnarray*} A \implies (B \implies C)& \mbox{ versus } & (A \implies B) \implies C\\ 0 \implies (0 \implies 0)& \mbox{ versus } & (0 \implies 0) \implies 0\\ 0 \implies 1 & \mbox{ versus } & 1 \implies 0\\ 1 & \mbox{ versus } & 0 \\ \end{eqnarray*} which shows that they are not equivalent.