Rule of Association/Disjunction/Formulation 2/Reverse Implication
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Theorem
- $\vdash \paren {p \lor \paren {q \lor r} } \impliedby \paren {\paren {p \lor q} \lor r}$
Proof
By definition of $\impliedby$, we prove:
- $\vdash \paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }$
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 | $r \implies \paren {q \lor r}$ | Axiom $\text A 2$ | $r / q, q / p$ | |||
2 | $\paren {r \implies \paren {q \lor r} } \implies \paren {\paren {\paren {p \lor q} \lor r} \implies \paren {\paren {p \lor q} \lor \paren {q \lor r} } }$ | Axiom $\text A 4$ | $r / q, \paren {p \lor q} \,/\, p, \paren {q \lor r} \,/\, r$ | |||
3 | $\paren {\paren {p \lor q} \lor r} \implies \paren {\paren {p \lor q} \lor \paren {q \lor r} }$ | Rule $\text {RST} 3$ | 1, 2 | |||
4 | $q \implies \paren {r \lor q}$ | Axiom $\text A 2$ | $r / p$ | |||
5 | $\paren {q \implies \paren {r \lor q} } \implies \paren {\paren {p \lor q} \implies \paren {p \lor \paren {r \lor q} } }$ | Axiom $\text A 4$ | $\paren {r \lor q} \,/\, r$ | |||
6 | $\paren {p \lor q} \implies \paren {p \lor \paren {r \lor q} }$ | Rule $\text {RST} 3$ | 4, 5 | |||
7 | $\paren {p \lor \paren {r \lor q} } \implies \paren {\paren {r \lor q} \lor p}$ | Axiom $\text A 3$ | $\paren {r \lor q} \,/\, q$ | |||
8 | $\paren {p \lor q} \implies \paren {\paren {r \lor q} \lor p}$ | Hypothetical Syllogism | 6, 7 | |||
9 | $q \implies \paren {q \lor p}$ | Rule of Addition | $p \,/\, q, q \,/\, p$ | |||
10 | $q \implies \paren {q \lor s}$ | Rule of Addition | $s \,/\, q, q \,/\, p$ | |||
11 | $\paren {q \implies \paren {q \lor s} } \implies \paren {\paren {p \lor q} \implies \paren {p \lor \paren {q \lor s} } }$ | Axiom $\text A 4$ | $\paren {q \lor s} \,/\, r$ | |||
12 | $\paren {p \lor q} \implies \paren {p \lor \paren {q \lor s} }$ | Rule $\text {RST} 3$ | 10, 11 | |||
13 | $\paren {\paren {p \lor q} \implies \paren {p \lor \paren {q \lor s} } } \implies \paren {\paren {r \lor \paren {p \lor q} } \implies \paren {r \lor \paren {p \lor \paren {q \lor s} } } }$ | Axiom $\text A 4$ | $r \,/\, p, \paren {p \lor q} \,/\, q, \paren {p \lor \paren {q \lor s} } \,/\, r$ | |||
14 | $\paren {r \lor \paren {p \lor q} } \implies \paren {r \lor \paren {p \lor \paren {q \lor s} } }$ | Rule $\text {RST} 3$ | 12, 13 | |||
15 | $\paren {\paren {p \lor q} \lor \paren {q \lor r} } \implies \paren {\paren {p \lor \paren {q \lor r} } \lor \paren {p \lor q} }$ | Rule $\text {RST} 1$ | 8 | $\paren {p \lor q} \,/\, p, \paren {q \lor r} \,/\, q, p \,/\, r$ | ||
16 | $\paren {\paren {p \lor \paren {q \lor r} } \lor \paren {p \lor q} } \implies \paren {\paren {p \lor \paren {q \lor r} } \lor \paren {p \lor \paren {q \lor r} } }$ | Rule $\text {RST} 1$ | 14 | $\paren {p \lor \paren {q \lor r} } \,/\, r, r \,/\, s$ | ||
17 | $\paren {\paren {p \lor \paren {q \lor r} } \lor \paren {p \lor \paren {q \lor r} } } \implies \paren {p \lor \paren {q \lor r} }$ | Axiom $\text A 1$ | $\paren {p \lor \paren {q \lor r} } \,/\, p$ | |||
18 | $\paren {\paren {p \lor q} \lor \paren {q \lor r} } \implies \paren {\paren {p \lor \paren {q \lor r} } \lor \paren {p \lor \paren {q \lor r} } }$ | Hypothetical Syllogism | 15, 16 | |||
19 | $\paren {\paren {p \lor q} \lor \paren {q \lor r} } \implies \paren {p \lor \paren {q \lor r} }$ | Hypothetical Syllogism | 17, 18 | |||
20 | $\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }$ | Hypothetical Syllogism | 3, 19 |
$\blacksquare$
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: $D \, 5$ - $D \, 9$