Law of Excluded Middle implies Peirce's Law

Theorem

From the Law of Excluded Middle follows Peirce's Law:

$\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p$

Proof

By the tableau method of natural deduction:

$\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \lor \neg p$	Premise	(None)
2		2	$p$	Assumption	(None)
3		2	$\paren {\paren {p \implies q} \implies p} \implies p$	Sequent Introduction	2	True Statement is implied by Every Statement
4		4	$\neg p$	Assumption	(None)
5		4	$p \implies q$	Sequent Introduction	4	False Statement implies Every Statement
6		6	$\paren {p \implies q} \implies p$	Assumption	(None)
7		6, 5	$p$	Modus Ponendo Ponens: $\implies \mathcal E$	6, 5
8		5	$\paren {\paren {p \implies q} \implies p} \implies p$	Rule of Implication: $\implies \II$	6 – 7	Assumption 6 has been discharged
9		1	$\paren {\paren {p \implies q} \implies p} \implies p$	Proof by Cases: $\text{PBC}$	1, 2 – 3, 4 – 8	Assumptions 2 and 4 have been discharged

$\blacksquare$