Peirce's Law

Comment
A non-obvious result is that that Peirce's Law has the same strength as the Law of the Excluded Middle:

As shown in Peirce's Law/Formulation 1/Proof 1, Peirce's Law follows from Law of Excluded Middle.

As shown in Clavius's Law/Formulation 1/Proof 2 and Law of Excluded Middle/Sequent Form/Proof 3, Law of Excluded Middle follows from Peirce's Law.

Historical Note
Here is Peirce's own statement and proof of the law:


 * A fifth icon is required for the principle of excluded middle and other propositions connected with it. One of the simplest formulae of this kind is:


 * $\left({\left({x \, \mathop {-\!\!\!<} y}\right) \, \mathop {-\!\!\!<} x}\right) \, \mathop {-\!\!\!<} x$


 * This is hardly axiomatical. That it is true appears as follows.  It can only be false by the final consequent $x$ being false while its antecedent $\left({x \mathop {-\!\!\!<} y}\right) \mathop {-\!\!\!<} x$ is true.  If this is true, either its consequent, $x$ is true, when the whole formula would be true, or its antecedent $x \,-\!\!\!< y$ is false.  But in the last case the antecedent of $x \,-\!\!\!< y,$ that is $x,\!$ must be true.

(Peirce, CP 3.384).

Peirce goes on to point out an immediate application of the law:


 * From the formula just given, we at once get:


 * $\left({\left({x \, \mathop {-\!\!\!<} y}\right) \, \mathop {-\!\!\!<} a}\right) \, \mathop {-\!\!\!<} x$


 * where the $a$ is used in such a sense that $\left({x \, \mathop {-\!\!\!<} y}\right) \, \mathop {-\!\!\!<} a$ means that from $\left({x \, \mathop {-\!\!\!<} y}\right)$ every proposition follows. With that understanding, the formula states the principle of excluded middle, that from the falsity of the denial of $x$ follows the truth of $x$.

(Peirce, CP 3.384).

Peirce uses the sign of illation $-\!\!\!<$ for implication. In one place he explains $-\!\!\!<$ as a variant of the sign $\le$ for less than or equal to; in another place he suggests that $A \mathop {-\!\!\!<} B$ is an iconic way of representing a state of affairs where $A$ in every way that it can be, is $B$.