Rule of Association/Disjunction/Formulation 1

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Theorem

$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$


Proof 1

By the tableau method of natural deduction:

$p \lor \paren {q \lor r} \vdash \paren {p \lor q} \lor r$
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:

$\paren {p \lor q} \lor r \vdash p \lor \paren {q \lor r} $
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.

$\begin{array}{|ccccc||ccccc|} \hline p & \lor & (q & \lor & r) & (p & \lor & q) & \lor & r \\ \hline \F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\ \F & \T & \F & \T & \T & \F & \F & \F & \T & \T \\ \F & \T & \T & \T & \F & \F & \T & \T & \T & \F \\ \F & \T & \T & \T & \T & \F & \T & \T & \T & \T \\ \T & \T & \F & \F & \F & \T & \T & \F & \T & \F \\ \T & \T & \F & \T & \T & \T & \T & \F & \T & \T \\ \T & \T & \T & \T & \F & \T & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$


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