The person next to me in tutorial asked if associativity worked between conjunction and disjunction. For example:
\[ P \wedge (Q \vee R) \overset{?}{\Leftrightarrow} (P \wedge Q) \vee R \]At first, I said "sure, why not?" not really thinking about it, but really, the two expressions mean different things.
\[ P \wedge (Q \vee R) \Leftrightarrow (P \wedge Q) \vee (P \wedge R) \\
(P \wedge Q) \vee R \Leftrightarrow (P \vee R) \wedge (Q \vee R)\]Namely, the second one is true whenever \(R\) is true. As a truth table,
\[\begin{array}{c c c | c | c}
P & Q & R & P \wedge (Q \vee R) & (P \wedge Q) \vee R \\ \hline
T & T & T & T & T \\
T & T & F & T & T \\
T & F & T & T & T \\
T & F & F & F & F \\
F & T & T & F & T \\
F & T & F & F & F \\
F & F & T & F & T \\
F & F & F & F & F \\
\end{array}\]Good to know.
\[ P \wedge (Q \vee R) \overset{?}{\Leftrightarrow} (P \wedge Q) \vee R \]At first, I said "sure, why not?" not really thinking about it, but really, the two expressions mean different things.
\[ P \wedge (Q \vee R) \Leftrightarrow (P \wedge Q) \vee (P \wedge R) \\
(P \wedge Q) \vee R \Leftrightarrow (P \vee R) \wedge (Q \vee R)\]Namely, the second one is true whenever \(R\) is true. As a truth table,
\[\begin{array}{c c c | c | c}
P & Q & R & P \wedge (Q \vee R) & (P \wedge Q) \vee R \\ \hline
T & T & T & T & T \\
T & T & F & T & T \\
T & F & T & T & T \\
T & F & F & F & F \\
F & T & T & F & T \\
F & T & F & F & F \\
F & F & T & F & T \\
F & F & F & F & F \\
\end{array}\]Good to know.
Hey! Don't just dismiss a question with a 'sure, why not'. There are many learning styles, and you might very well have figured yours out, but if you haven't tried all of them, one of the best ways to really 'learn' is to explain to others.
ReplyDeleteGreat Slog so far, keep up the good work! Work through some problems that Prof. Heap suggested, or any others you find online, to really put the stuff you're learning in this course into a different perspective.