Conjunction implies Disjunction/Proof 2
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Theorem
- $\vdash \paren {p \land q} \implies \paren {p \lor q}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \land q$ | Assumption | (None) | ||
| 2 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 3 | 1 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 2 | ||
| 4 | $\paren {p \land q} \implies \paren {p \lor q}$ | Rule of Implication: $\implies \II$ | 1 – 3 | Assumption 1 has been discharged |
$\blacksquare$