# Proof by Cases with Contradiction

## Theorem

$\vdash p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$

## Proof

By the tableau method of natural deduction:

$\vdash p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Assumption (None)
2 1 $p \lor q$ Rule of Addition: $\lor \II_1$ 1
3 1 $p \lor \neg q$ Rule of Addition: $\lor \II_2$ 1
4 1 $\left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ Rule of Conjunction: $\land \II$ 2, 3
5 $p \implies \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ Rule of Implication: $\implies \II$ 1 – 4 Assumption 1 has been discharged
6 6 $\left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ Assumption (None)
7 6 $p \lor \left({q \land \neg q}\right)$ Sequent Introduction 6 Disjunction Distributes over Conjunction
8 8 $p$ Assumption (None)
9 $\neg \left({q \land \neg q}\right)$ Theorem Introduction (None) Principle of Non-Contradiction: Formulation 2
10 6 $p$ Modus Tollendo Ponens $\mathrm {MTP}_2$ 7, 9
11 6 $p$ Proof by Cases: $\text{PBC}$ 6, 8 – 8, 9 – 10 Assumptions 8 and 9 have been discharged
12 $\left({p \lor q}\right) \land \left({p \lor \neg q}\right) \implies p$ Rule of Implication: $\implies \II$ 6 – 11 Assumption 6 has been discharged
13 $p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$ Biconditional Introduction: $\iff \II$ 5, 12

$\blacksquare$