# Conjunction with Law of Excluded Middle

## Theorem

$\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q}$

## Proof

By the tableau method of natural deduction:

$\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q}$
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Assumption (None)
2 $q \lor \neg q$ Law of Excluded Middle (None)
3 3 $q$ Assumption (None)
4 1, 3 $p \land q$ Rule of Conjunction: $\land \II$ 1, 2
5 1, 3 $\paren {p \land q} \lor \paren {p \land \neg q}$ Rule of Addition: $\lor \II_1$ 4
6 6 $\neg q$ Assumption (None)
7 1, 6 $p \land \neg q$ Rule of Conjunction: $\land \II$ 1, 6
8 1, 6 $\paren {p \land q} \lor \paren {p \land \neg q}$ Rule of Addition: $\lor \II_2$ 7
9 1 $\paren {p \land q} \lor \paren {p \land \neg q}$ Proof by Cases: $\text{PBC}$ 2, 3 – 5, 6 – 8 Assumptions 3 and 6 have been discharged
10 $p \implies \paren {p \land q} \lor \paren {p \land \neg q}$ Rule of Implication: $\implies \II$ 1 – 9 Assumption 1 has been discharged
11 11 $\paren {p \land q} \lor \paren {p \land \neg q}$ Assumption (None)
12 11 $p \land \paren {q \lor \neg q}$ Sequent Introduction 11 Conjunction Distributes over Disjunction
13 11 $p$ Rule of Simplification: $\land \EE_2$ 11
14 $\paren {p \land q} \lor \paren {p \land \neg q} \implies p$ Rule of Implication: $\implies \II$ 11 – 13 Assumption 11 has been discharged
15 $p \iff \paren {p \land q} \lor \paren {p \land \neg q}$ Biconditional Introduction: $\iff \II$ 12, 14

$\blacksquare$