# Law of Excluded Middle/Sequent Form

## Theorem

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

$\vdash p \lor \neg p$

## Proof 1

By the tableau method of natural deduction:

$\vdash p \lor \neg p$
Line Pool Formula Rule Depends upon Notes
1 $p \lor \neg p$ Law of Excluded Middle (None)

$\blacksquare$

## Proof 2

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 $\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ Theorem Introduction (None) Hypothetical Syllogism
2 $\paren {\paren {p \lor p} \implies p} \implies \paren {\paren {p \implies \paren {p \lor p} } \implies \paren {p \implies p} }$ Rule $\text {RST} 1$ 1 $p \lor p \, / \, q$, $p \, / \, r$
3 $\paren {p \lor p} \implies p$ Axiom $\text A 1$
4 $\paren {p \implies \paren {p \lor p} } \implies \paren {p \implies p}$ Rule $\text {RST} 3$ 2, 3
5 $p \implies \paren {p \lor p}$ Axiom $\text A 2$, Rule $\text {RST} 1$ $p \, / \, q$
6 $p \implies p$ Rule $\text {RST} 3$ 4, 5
7 $\neg p \lor p$ Rule $\text {RST} 2 \, (2)$ 6
8 $\paren {p \lor q} \implies \paren {q \lor p}$ Axiom $\text A 3$
9 $\paren {\neg p \lor p} \implies \paren {p \lor \neg p}$ Rule $\text {RST} 1$ 8 $p \, / \, q$, $\neg p \, / \, p$
10 $p \lor \neg p$ Rule $\text {RST} 3$ 7, 9

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables to the proposition $\vdash p \lor \neg p$.

As can be seen by inspection, the truth value of the main connective, that is $\lor$, is $\T$ for each boolean interpretation for $p$.

$\begin{array}{|cccc|} \hline p & \lor & \neg & p \\ \hline \F & \T & \T & \F \\ \T & \T & \F & \T \\ \hline \end{array}$

$\blacksquare$