Rule of Addition/Sequent Form/Formulation 2/Proof 1
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Theorem
| \(\text {(1)}: \quad\) | \(\ds \vdash p\) | \(\implies\) | \(\ds \paren {p \lor q}\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \vdash q\) | \(\implies\) | \(\ds \paren {p \lor q}\) |
Proof
Form 1
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p$ | Premise | (None) | ||
| 2 | 1 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 1 | ||
| 3 | $p \implies \paren {p \lor q}$ | Rule of Implication: $\implies \II$ | 1 – 3 | Assumption 1 has been discharged |
$\blacksquare$
Form 2
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $q$ | Premise | (None) | ||
| 2 | 1 | $p \lor q$ | Rule of Addition: $\lor \II_2$ | 1 | ||
| 3 | $q \implies \paren {p \lor q}$ | Rule of Implication: $\implies \II$ | 1 – 3 | Assumption 1 has been discharged |
$\blacksquare$