Rule of Association/Conjunction/Formulation 1/Proof 1
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Theorem
- $p \land \left({q \land r}\right) \dashv \vdash \left({p \land q}\right) \land r$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \land \left({q \land r}\right)$ | Premise | (None) | ||
2 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $q \land r$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
4 | 1 | $q$ | Rule of Simplification: $\land \EE_1$ | 3 | ||
5 | 1 | $r$ | Rule of Simplification: $\land \EE_2$ | 3 | ||
6 | 1 | $p \land q$ | Rule of Conjunction: $\land \II$ | 2, 4 | ||
7 | 1 | $\left({p \land q}\right) \land r$ | Rule of Conjunction: $\land \II$ | 6, 5 |
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\left({p \land q}\right) \land r$ | Premise | (None) | ||
2 | 1 | $p \land q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $r$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
4 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 2 | ||
5 | 1 | $q$ | Rule of Simplification: $\land \EE_2$ | 2 | ||
6 | 1 | $q \land r$ | Rule of Conjunction: $\land \II$ | 5, 3 | ||
7 | 1 | $p \land \left({q \land r}\right)$ | Rule of Conjunction: $\land \II$ | 4, 6 |
$\blacksquare$