I want to prove $\Diamond (P \lor Q) \Rightarrow \Diamond P \lor \Diamond Q$ It was a biconditional, but I have proved the other one. Thanks for the answer. Please use Garson's method. Thanks. I am stuck on this, any help. Please.
I don't have any work to show for it because I don't know how to proceed.
Maybe try proof by contradiction
Here is one possible approach $\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$ \begin{align} \fitch{\Diamond(p\lor q)}{ \neg\square\neg(p\lor q)\hspace{7.8ex}\Diamond\text{Def}\\ \fitch{\neg(\Diamond p\lor\Diamond q)} {\neg\Diamond p\land\neg\Diamond q\hspace{5ex}\text{DM}\\ \neg\Diamond p\hspace{12ex}\&\text{Out}\\ \fitch{\neg\square\neg p}{\Diamond p\hspace{10ex}\Diamond\text{Def}\\ \bot\hspace{11ex}\bot\text{In}}\\\square \neg p\hspace{13.5ex}\text{IP}\\ \neg\Diamond q\hspace{12ex}\&\text{Out}\\ \fitch{\neg\square\neg q}{\Diamond q\hspace{10ex}\Diamond\text{Def}\\ \bot\hspace{11ex}\bot\text{In}}\\\square\neg q\hspace{13.5ex}\text{IP}\\ \fitch{\square} {\neg p\hspace{10.2ex}\square\text{Out}\\ \neg q\hspace{10.2ex}\square\text{Out}\\ \fitch{p\lor q}{\fitch{p}{\bot}\fitch{q}{\bot}\\\bot\hspace{7ex}\lor\text{Out}}\\\neg(p\lor q)\hspace{5ex}\text{IP}} \\\square\neg(p\lor q)\hspace{6.5ex}\square\text{In}\\ \bot\hspace{14.8ex}\bot\text{In}}\\ \Diamond p\lor\Diamond q\hspace{12.4ex}\text{IP}} \end{align}