Rule of Association/Disjunction/Formulation 2/Forward Implication
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Theorem
- $\vdash \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$
Proof
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 | $\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }$ | Rule of Association: Reverse Implication | ||||
| 2 | $\paren {q \lor r} \implies \paren {r \lor q}$ | Axiom $\text A 3$ | $q \,/\, p, r \,/\, q$ | |||
| 3 | $\paren {\paren {q \lor r} \implies \paren {r \lor q} } \implies \paren {\paren {p \lor \paren {q \lor r} } \implies \paren {p \lor \paren {r \lor q} } }$ | Axiom $\text A 4$ | $\paren {q \lor r} \,/\, q, \paren {r \lor q} \,/\, r$ | |||
| 4 | $\paren {p \lor \paren {q \lor r} } \implies \paren {p \lor \paren {r \lor q} }$ | Rule $\text {RST} 3$ | 2, 3 | |||
| 5 | $\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {r \lor q} }$ | Hypothetical Syllogism | 1, 4 | |||
| 6 | $\paren {p \lor \paren {r \lor q} } \implies \paren {\paren {r \lor q} \lor p}$ | Axiom $\text A 3$ | $\paren {r \lor q} \,/\, q$ | |||
| 7 | $\paren {\paren {p \lor q} \lor r} \implies \paren {\paren {r \lor q} \lor p}$ | Hypothetical Syllogism | 5, 6 | |||
| 8 | $\paren {r \lor \paren {p \lor q} } \implies \paren {\paren {p \lor q} \lor r}$ | Axiom $\text A 3$ | $r \,/\, p, \paren {p \lor q} \,/\, q$ | |||
| 9 | $\paren {r \lor \paren {p \lor q} } \implies \paren {\paren {r \lor q} \lor p}$ | Hypothetical Syllogism | 7, 8 | |||
| 10 | $\paren {q \lor p} \implies \paren {p \lor q}$ | Axiom $\text A 3$ | $p / q, q / p$ | |||
| 11 | $\paren {\paren {q \lor p} \implies \paren {p \lor q} } \implies \paren {\paren {r \lor \paren {q \lor p} } \implies \paren {r \lor \paren {p \lor q} } }$ | Axiom $\text A 4$ | $r \,/\, p, \paren {q \lor p} \,/\, q, \paren {p \lor q} \,/\, r$ | |||
| 12 | $\paren {r \lor \paren {q \lor p} } \implies \paren {r \lor \paren {p \lor q} }$ | Rule $\text {RST} 3$ | 10, 11 | |||
| 13 | $\paren {r \lor \paren {q \lor p} } \implies \paren {\paren {r \lor q} \lor p}$ | Hypothetical Syllogism | 9, 12 | |||
| 14 | $\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$ | Rule $\text {RST} 1$ | 13 | $r / p, p / r$ |
$\blacksquare$
Sources
- 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 4.7$: The Derivation of Formulae: $D \, 10$