Proving equivalence of $P$ and $P \wedge (P \vee Q)$ How would you prove the equivalence of
$P$ and $P \wedge (P \vee Q)$
What I have so far is:
$P \iff P \vee (Q \wedge \neg Q)$ (identity)
$\iff (P \vee Q) \wedge (P \vee \neg Q)$ (distributivity)
$\iff ((P \wedge (P \vee Q)) \vee  \neg Q \wedge (P \vee Q))$ (distributivity)
$\iff ((P \wedge (P \vee Q)) \vee (\neg Q \wedge P))$ (identity)
The term appears on the right but there is also something extra. 
 A: Let $T$ represent a tautology and $C$ represent a contradiction. Then here's one way to prove equivalence:
$$ \begin{array}{rcll}
P \land (P \lor Q) &\iff& (P \lor P) \land (P \lor Q) & \quad\text{by Idempotent Law}\\
 &\iff& P \lor (P \land  Q) & \quad\text{by Distributive Law}\\
 &\iff& (P \land T) \lor (P \land  Q) & \quad\text{by Identity Law}\\
 &\iff& P \land (T \lor Q) & \quad\text{by Distributive Law}\\
 &\iff& P \land T & \quad\text{by Domination Law}\\
 &\iff& P & \quad\text{by Identity Law}\\
\end{array}$$
as desired. The identity:

$$P \land (P \lor Q) \iff P \iff P \lor (P \land  Q)$$

is known as Absorption Law. A shorter proof is as follows:
$$ \begin{array}{rcll}
P \land (P \lor Q) &\iff& (P \lor C) \land (P \lor Q) & \quad\text{by Identity Law}\\
 &\iff& P \lor (C \land  Q) & \quad\text{by Distributive Law}\\
 &\iff& P \lor C & \quad\text{by Domination Law}\\
 &\iff& P & \quad\text{by Identity Law}\\
\end{array}$$
A: Or, considering the Implicational propositional calculus, you can overcome this simple deduction:
$$P\wedge(P\vee Q)/\therefore P$$
and $$P\wedge\sim(P\wedge(P\vee Q))/\therefore P\wedge(\sim P\vee(\sim P\wedge \sim Q))/\therefore( \sim P\vee(\sim P\wedge \sim Q))\\\\ \therefore( \sim P\vee\sim P) \wedge (\sim P\vee\sim Q)\therefore\sim P$$ which is a contradiction, so both statements are logically equivalent.
A: Suppose p=0 (falsity).  Then (P∧(P∨Q))=(0$\land$(0$\lor$ q))=0 since (0$\land$x)=0 for all x.
Suppose p=1.  Then (P∧(P∨Q))=(1$\land$(1 $\lor$ q))=(1 $\land$ 1)=1, since (1$\lor$x)=1 for all x.
Since p belongs to {0, 1} this completes the demonstration.
A: There are many ways to prove this, depending on what laws you're allowing yourself to use.  Personally I don't like 'both directions separately' arguments.  Writing down the statement to be proved as
$$
P \equiv P \land (P \lor Q)
$$
(where $\;\equiv\;$ is an alternative notation for $\;\iff\;$), I would just try to simplify this statement as much as possible, e.g., as follows:
\begin{align}
& P \equiv P \land (P \lor Q) \\
\equiv & \;\;\;\;\;\text{"$\;P \equiv P \land \dots\;$ is one of the many ways to rewrite $\;P \Rightarrow \dots\;$"} \\
& P \Rightarrow P \lor Q \\
\equiv & \;\;\;\;\;\text{"$\;\lnot P \lor \dots\;$ is the most common way to rewrite $\;P \Rightarrow \dots\;$"} \\
& \lnot P \lor P \lor Q \\
\equiv & \;\;\;\;\;\text{"excluded middle"} \\
& \text{true} \lor Q \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& \text{true} \\
\end{align}
This completes the proof.
Update. For an even shorter proof, one can use (what Dijkstra and Scholten on p. 37 call) the golden rule: $$R \equiv R \land S \equiv R \lor S \equiv S$$  This can be applied in a lot of different ways, since $\;\equiv\;$ is both associative an symmetric.  Here we recognize that the statement to be proved has the same shape as part of this rule, so that we can use it as follows:
\begin{align}
& P \equiv P \land (P \lor Q) \\
\equiv & \;\;\;\;\;\text{"rewrite using the golden rule -- to replace $\;\land\;$ by $\;\lor\;$ "} \\
& P \lor P \lor Q \equiv P \lor Q \\
\equiv & \;\;\;\;\;\text{"simplify $\;P \lor P\;$ to $\;P\;$ in left hand side; identity"} \\
& \text{true} \\
\end{align}
A: Adriano's proof works fine, but to establish the equivalence $P$ and $P \land (P \lor Q)$ why not take the two directions of the biconditional separately? One direction is just trivial, and the other almost so.
