Peirce's Law is Equivalent to Law of Excluded Middle

Theorem
Peirce's Law:
 * $\left({p \implies q}\right) \implies p \vdash p$

is logically equivalent to the Law of the Excluded Middle:
 * $\vdash p \lor \neg p$

That is, Peirce's Law holds if and only if the Law of the Excluded Middle holds.

Proof
Let $PL$ stand for the proposition Peirce's Law holds.

Let $LEM$ stand for the proposition The Law of the Excluded Middle holds.

Sufficient Condition
First we show that $LEM \implies PL$.

Let us assume that $LEM$ is true.

Then:

Suppose $\neg PL$:
 * $\left({p \implies q}\right) \implies p \not \vdash p$

Then there exists a model $\mathcal M: \left\{{p, q}\right\} \to \left\{{T, F}\right\}$ such that:

But then:
 * $\left({F \implies q}\right) \implies F$

That is:
 * $F \implies q \vdash F$

From Implication Properties:
 * $F \implies q \vdash T$

From this contradiction we see that it can not be the case that $\left({p \implies q}\right) \implies p \not \vdash p$ must be false.

Therefore $PL$ is true:
 * $\left({p \implies q}\right) \implies p \vdash p$

that is, Peirce's Law holds.

So $LEM \implies PL$.

Necessary Condition
Now assume that Peirce's Law holds.

Suppose that $p$ is not false.

If we can show that $p$ therefore has to be true, we have proved that $LEM$ is true.

If $p \implies q$ is true, it must be that $p$ is true.

But when $q$ is false, $p \implies q$ does not hold when $p$ is true.

Therefore $\left({p \implies q}\right) \implies p$ is true.

But if $PL$ is true, that means $p$ is true.

Thus we see that $PL \implies LEM$.