Rule of Association/Disjunction/Formulation 2
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Theorem
- $\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$
Forward Implication
- $\vdash \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$
Reverse Implication
- $\vdash \paren {p \lor \paren {q \lor r} } \impliedby \paren {\paren {p \lor q} \lor r}$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \lor \paren {q \lor r}$ | Assumption | (None) | ||
2 | 1 | $\paren {p \lor q} \lor r$ | Sequent Introduction | 1 | Rule of Association: Formulation 1 | |
3 | $\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | ||
4 | 4 | $\paren {p \lor q} \lor r$ | Assumption | (None) | ||
5 | 4 | $p \lor \paren {q \lor r}$ | Sequent Introduction | 4 | Rule of Association: Formulation 1 | |
6 | $\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | ||
7 | $\paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$ | Biconditional Introduction: $\iff \II$ | 3, 6 |
$\blacksquare$
Proof 2
This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.
By the tableau method:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$ | Rule of Association: Forward Implication | ||||
2 | $\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }$ | Rule of Association: Reverse Implication | ||||
3 | $\paren {\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r} } \land \paren {\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} } }$ | Rule $\text {RST} 4$ | 1, 2 | |||
4 | $\paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$ | Rule $\text {RST} 2 (3)$ | 3 |
$\blacksquare$
Sources
- 1946: Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences (2nd ed.) ... (previous) ... (next): $\S \text{II}.13$: Symbolism of sentential calculus
- 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 3.6$: Reference Formulae: $RF \, 11$
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T54}$
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $3$: The Method of Deduction: $3.2$: The Rule of Replacement: $12.$
- 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{II}$: The Logic of Statements $(2): \ 6$: Using logical equivalences: $11$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(1) \ \text{(vi)}$
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.14$: Exercise $12 \ (3)$