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 $$PL \implies LEM$$.