Constructive Dilemma/Formulation 1/Proof 1
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Theorem
\(\ds p \implies q\) | \(\) | \(\ds \) | ||||||||||||
\(\ds r \implies s\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds p \lor r \implies q \lor s\) | \(\) | \(\ds \) |
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies q$ | Premise | (None) | ||
2 | 2 | $r \implies s$ | Premise | (None) | ||
3 | 3 | $p \lor r$ | Assumption | (None) | ||
4 | 4 | $p$ | Assumption | (None) | ||
5 | 1, 4 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 4 | ||
6 | 1, 4 | $q \lor s$ | Rule of Addition: $\lor \II_1$ | 5 | ||
7 | 7 | $r$ | Assumption | (None) | ||
8 | 2, 7 | $s$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 7 | ||
9 | 2, 7 | $q \lor s$ | Rule of Addition: $\lor \II_2$ | 8 | ||
10 | 1, 2, 3 | $q \lor s$ | Proof by Cases: $\text{PBC}$ | 3, 4 – 6, 7 – 9 | Assumptions 4 and 7 have been discharged | |
11 | 1, 2 | $p \lor r \implies q \lor s$ | Rule of Implication: $\implies \II$ | 3 – 10 | Assumption 3 has been discharged |
$\blacksquare$