# Law of Excluded Middle/Sequent Form/Proof 3

## Theorem

The Law of Excluded Middle can be symbolised by the sequent:

$\vdash p \lor \neg p$

## Proof

This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.

By the tableau method:

$\vdash p \lor \neg p$
Line Pool Formula Rule Depends upon Notes
1 $\left({q \implies r}\right) \implies \left({\left({p \implies q}\right) \implies \left({p \implies r}\right)}\right)$ Theorem Introduction (None) Hypothetical Syllogism
2 $\left({\left({p \lor p}\right) \implies p}\right) \implies \left({\left({p \implies \left({p \lor p}\right)}\right) \implies \left({p \implies p}\right)}\right)$ Rule $RST \, 1$ 1 $p \lor p \, / \, q$, $p \, / \, r$
3 $(p \lor p) \implies p$ Axiom $A1$
4 $\left({p \implies \left({p \lor p}\right)}\right) \implies \left({p \implies p}\right)$ Rule $RST \, 3$ 2, 3
5 $p \implies (p \lor p)$ Axiom $A2$, Rule $RST \, 1$ $p \, / \, q$
6 $p \implies p$ Rule $RST \, 3$ 4, 5
7 $\neg p \lor p$ Rule $RST \, 2 \, (2)$ 6
8 $(p \lor q) \implies (q \lor p)$ Axiom $A3$
9 $(\neg p \lor p) \implies (p \lor \neg p)$ Rule $RST \, 1$ 8 $p \, / \, q$, $\neg p \, / \, p$
10 $p \lor \neg p$ Rule $RST \, 3$ 7, 9

$\blacksquare$