Proof in logic that $P \Leftrightarrow Q$ is the same as $ (P \lor Q) \rightarrow (P \land Q)$

Question

How is it possible to prove that $P \Leftrightarrow Q$ is the same as $ (P \lor Q) \rightarrow (P \land Q)$ using logic laws?

Answer 1

First, you can rewrite $(P∨Q)→(P∧Q)$ as $\lnot (P∨Q)∨(P∧Q)$.

Then use De Morgan's laws, i.e. : $¬(P∨Q)$ is equivalent to $(¬P∧¬Q)$ to get :

$(¬P∧¬Q)∨(P∧Q)$.

Then you need the Ditributive property of truth functional connectives to transform the last formula into :

$[(¬P∧¬Q) \lor P] \land [(¬P∧¬Q) \lor Q]$

and applaing it again :

$(\lnot P \lor P ) \land (\lnot Q \lor P) \land (\lnot P \lor Q) \land (\lnot Q \lor Q)$.

Finally, we use the Identity laws : $\lnot P \lor P \equiv T$ and $T \land P \equiv P$, to simplify the formula and get :

$(\lnot Q \lor P) \land (\lnot P \lor Q)$.

But this is : $P \leftrightarrow Q$.

Note

You can find a useful summary of the main concepts and of the laws involved in the proof above into this presentation.

Answer 2

You can prove that $P\Rightarrow Q, Q\Rightarrow P \dashv\vdash (P\lor Q)\Rightarrow (P\land Q)$ by showing first that $P\Rightarrow Q, Q\Rightarrow P \vdash (P\lor Q) \Rightarrow (P\land Q)$ and then that $(P\lor Q) \Rightarrow (P\land Q)\vdash (P\Rightarrow Q) \land (Q\Rightarrow P)$: