Rule of Association/Disjunction/Formulation 1
Theorem
- $p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \lor \paren {q \lor r}$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | By assuming the first main disjunct ... | |
3 | 2 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 2 | ||
4 | 2 | $\paren {p \lor q} \lor r$ | Rule of Addition: $\lor \II_1$ | 3 | ... the conclusion is derived | |
5 | 5 | $q \lor r$ | Assumption | (None) | Then assume the second main disjunct ... | |
6 | 6 | $q$ | Assumption | (None) | ... and by assuming the first disjunct of that second main disjunct ... | |
7 | 6 | $p \lor q$ | Rule of Addition: $\lor \II_2$ | 6 | ||
8 | 6 | $\paren {p \lor q} \lor r$ | Rule of Addition: $\lor \II_1$ | 7 | ... the conclusion is derived | |
9 | 9 | $r$ | Assumption | (None) | Then assume the second disjunct of that second main disjunct ... | |
10 | 9 | $\paren {p \lor q} \lor r$ | Rule of Addition: $\lor \II_2$ | 9 | ... and likewise the same conclusion is derived | |
11 | 5 | $\paren {p \lor q} \lor r$ | Proof by Cases: $\text{PBC}$ | 5, 6 – 8, 9 – 10 | Assumptions 6 and 9 have been discharged | |
12 | 1 | $\paren {p \lor q} \lor r$ | Proof by Cases: $\text{PBC}$ | 1, 2 – 4, 5 – 11 | Assumptions 2 and 5 have been discharged |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \lor q} \lor r$ | Premise | (None) | ||
2 | 2 | $p \lor q$ | Assumption | (None) | ||
3 | 3 | $p$ | Assumption | (None) | ||
4 | 3 | $p \lor \paren {q \lor r}$ | Rule of Addition: $\lor \II_1$ | 3 | ||
5 | 5 | $q$ | Assumption | (None) | ||
6 | 5 | $q \lor r$ | Rule of Addition: $\lor \II_1$ | 5 | ||
7 | 5 | $p \lor \paren {q \lor r}$ | Rule of Addition: $\lor \II_2$ | 6 | ||
8 | 2 | $p \lor \paren {q \lor r}$ | Proof by Cases: $\text{PBC}$ | 2, 3 – 4, 5 – 7 | Assumptions 3 and 5 have been discharged | |
9 | 9 | $r$ | Assumption | (None) | ||
10 | 9 | $q \lor r$ | Rule of Addition: $\lor \II_2$ | 9 | ||
11 | 9 | $p \lor \paren {q \lor r}$ | Rule of Addition: $\lor \II_2$ | 10 | ||
12 | 1 | $p \lor \paren {q \lor r}$ | Proof by Cases: $\text{PBC}$ | 1, 2 – 8, 9 – 11 | Assumptions 2 and 9 have been discharged |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline p & \lor & (q & \lor & r) & (p & \lor & q) & \lor & r \\ \hline \F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\ \F & \T & \F & \T & \T & \F & \F & \F & \T & \T \\ \F & \T & \T & \T & \F & \F & \T & \T & \T & \F \\ \F & \T & \T & \T & \T & \F & \T & \T & \T & \T \\ \T & \T & \F & \F & \F & \T & \T & \F & \T & \F \\ \T & \T & \F & \T & \T & \T & \T & \F & \T & \T \\ \T & \T & \T & \T & \F & \T & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Also see
Sources
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$