I am really confused about this and not sure how to show $P\iff Q$ with the ↓ arrow and only the ↓ arrow. I understand that $P \iff Q$ is $P\implies Q$ and $Q\implies P$.
I also know that $P\implies Q$ is also $\neg P$ or $Q$.
If anyone can help me represent $P\iff Q$ with only the ↓ arrow, I would really appreciate it.
First let's express $\neg, \lor$ and $\land$ using $\downarrow$, where $(P\downarrow Q) \iff (\neg P\land \neg Q)$: $$\begin{align} \neg P &:= (P\downarrow P) \\ (P\lor Q) &:= \neg(P\downarrow Q) \\ (P\land Q) &:= \neg(\neg P\lor \neg Q). \\ \end{align}$$ Now it's easy to define $\leftrightarrow$: $$ (P \leftrightarrow Q) := (P\land Q) \lor (\neg P \land \neg Q).\tag{$Equiv_1$} $$ An equivalent definition is: $$(P \leftrightarrow Q) := (\neg P\lor Q)\land(P\lor \neg Q).\tag{$Equiv_2$} $$
Eliminating the defined symbols $\neg, \lor$ and $\land$ in ($Equiv_1$) or ($Equiv_2$) is left as a mechanical exercise. (If you need to do so, it looks like ($Equiv_2$) is the easier definition to work with — it has only one $\land$ not two, so the resulting formula should be shorter than what you'd get starting from ($Equiv_1$).)
