= Solution
Expand three cases as two applications of <disjunction elimination rule>. For the association $(\phi\lor\psi)\lor\rho$, the first outer branch assumes $\phi\lor\psi$ and applies the given $\phi$ and $\psi$ subproofs to obtain $\kappa$. The other outer branch is the given $\rho$ subproof. In schematic <natural deduction> form:
``
1 (φ ∨ ψ) ∨ ρ given
2 | φ ∨ ψ first outer case
3 | | φ first inner case
4 | | κ given φ-subproof
5 | | ψ second inner case
6 | | κ given ψ-subproof
7 | κ ∨-elim, 2, cases 3–4 and 5–6
8 | ρ second outer case
9 | κ given ρ-subproof
10 κ ∨-elim, 1, cases 2–7 and 8–9
``
All three case assumptions are discharged and common surrounding assumptions remain. If the original three-way <logical disjunction> is parsed as $\phi\lor(\psi\lor\rho)$, use the $\phi$ subproof as the first outer branch and eliminate $\psi\lor\rho$ inside the second. Thus either binary association has the asserted <three-case disjunction elimination>, without presupposing an unproved <associativity>.
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