Disjunction of Conjunctions
Jump to navigation
Jump to search
Theorem
- $\left({p \land q}\right) \lor \left({r \land s}\right) \vdash p \lor r$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\left({p \land q}\right) \lor \left({r \land s}\right)$ | Premise | (None) | ||
2 | 2 | $p \land q$ | Assumption | (None) | ||
3 | 2 | $p$ | Rule of Simplification: $\land \EE_1$ | 2 | ||
4 | 2 | $p \lor r$ | Rule of Addition: $\lor \II_1$ | 3 | ||
5 | 5 | $r \land s$ | Assumption | (None) | ||
6 | 5 | $r$ | Rule of Simplification: $\land \EE_1$ | 5 | ||
7 | 5 | $p \lor r$ | Rule of Addition: $\lor \II_2$ | 6 | ||
8 | 1 | $p \lor r$ | Proof by Cases: $\text{PBC}$ | 1, 2 – 4, 5 – 7 | Assumptions 2 and 5 have been discharged |
$\blacksquare$