# Rule of Association/Disjunction/Formulation 2/Forward Implication

## 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:

$\vdash \paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$
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 $r / p, p / r$

$\blacksquare$