-(p ↓ q)⇐⇒(-p ↑-q)
I need to prove this but are not sure what the down and up arrows mean.
Do I have to construct a truth table?
Thanks,
Chris
So far as I know the definitions vary here. I can give you the one used by WVO Quine:
$\displaystyle \begin{array}{*{20}c} P &\vline & Q &\vline & {P \uparrow Q} \\\hline {T} &\vline & {T} &\vline & {F} \\ {T} &\vline & {F} &\vline & {T} \\ {F} &\vline & {T} &\vline & {T} \\ {F} &\vline & {F} &\vline & {T} \\ \end{array} $ and $\displaystyle \begin{array}{*{20}c} P &\vline & Q &\vline & {P \downarrow Q} \\\hline {T} &\vline & {T} &\vline & {F} \\ {T} &\vline & {F} &\vline & {F} \\ {F} &\vline & {T} &\vline & {F} \\ {F} &\vline & {F} &\vline & {T} \\ \end{array} $
I agree. The notations $\displaystyle \downarrow$ and $\displaystyle \uparrow$ probably refer to Peirce's arrow (NOR) and Sheffer stroke (NAND), respectively.
This depends on the assignment. There are many ways of proving a propositional formula, from constructing a truth table to giving a derivation in one of many formal systems.
Sorry, I also had to prove the formula ¬(p ↑ q)⇐⇒(¬p ↓¬q) so I added that information to the columns. since the columns for ¬(p ↓ q) equal the columns for (¬p ↑¬q) that formula is true. Since the columns for ¬(p ↑ q) equal the columns for (¬p ↓¬q) that formula is true.
Thank you for challanging me to finish thinking this problem through. I have suffered a recent brain injury and my thought process suffers from this.