Conjunction of Disjunction with Negation is Conjunction with Negation
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Theorem
- $\paren {p \lor q} \land \neg q \dashv \vdash p \land \neg q$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \land \neg q$ | Premise | (None) | ||
2 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 2 | ||
4 | 1 | $\neg q$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
5 | 1 | $\paren {p \lor q} \land \neg q$ | Rule of Conjunction: $\land \II$ | 3, 4 |
$\Box$
and its converse:
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \lor q} \land \neg q$ | Premise | (None) | ||
2 | 1 | $\paren {p \land \neg q} \lor \paren {q \land \neg q}$ | Sequent Introduction | 1 | Conjunction Distributes over Disjunction | |
3 | 3 | $q \land \neg q$ | Assumption | (None) | ||
4 | 3 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 3, 3 | ||
5 | $\neg \paren {q \land \neg q}$ | Proof by Contradiction: $\neg \II$ | 3 – 4 | Assumption 3 has been discharged | ||
6 | 1 | $p \land \neg q$ | Sequent Introduction | 2, 5 | Disjunctive Syllogism |
$\blacksquare$