I'm trying to calculate $(p\land q)\rightarrow (p \vee q)$ using just joint denial $\def\jd{\downarrow}\jd$. I'm not sure if I've done it right... any help would be greatly appreciated!
My workings: \begin{align*} (p \land q) \to (p \lor q) &\equiv (\neg p \jd \neg q) \to \neg (p \jd q)\\ &\equiv \bigl((p \jd p) \jd (q \jd q)\bigr) \to \bigl((p\jd q)\jd (p\jd q)\bigr)\\ &\equiv \neg \bigl((p \jd p) \jd (q \jd q)\bigr) \lor \bigl((p\jd q)\jd (p\jd q)\bigr)\\ &\equiv \bigl((p \jd p) \jd (q \jd q)\bigr)\jd \bigl((p \jd p) \jd (q \jd q)\bigr) \lor \bigl((p\jd q)\jd (p\jd q)\bigr)\\ &\equiv \neg \Bigl(\bigl((p \jd p) \jd (q \jd q)\bigr)\jd \bigl((p \jd p) \jd (q \jd q)\bigr) \jd \bigl((p\jd q)\jd (p\jd q)\bigr)\Bigr)\\ \end{align*}
You havent finished, there is still a $\neg$ in your term if $\neg t$ is your last term, you have to replace it with $t \downarrow t$, giving $$ \def\jd{\downarrow}\Bigl(\bigl((p \jd p) \jd (q \jd q)\bigr)\jd \bigl((p \jd p) \jd (q \jd q)\bigr) \jd \bigl((p\jd q)\jd (p\jd q)\bigr)\Bigr) \jd \Bigl(\bigl((p \jd p) \jd (q \jd q)\bigr)\jd \bigl((p \jd p) \jd (q \jd q)\bigr) \jd \bigl((p\jd q)\jd (p\jd q)\bigr)\Bigr) $$