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Proof step 7 from proof in-rm-v-in (previous step)

eq-entry(n, lref) ∈ (lref0 ≠ none ∧ lref0.vol = v ⊃ q0 - lref0.vol;eq-entry(m, lref0) ' + (q0 - v)), ∀ n, lref, v. eq-entry(n, lref) ∈ q0 - v → (∃ n. eq-entry(n, lref) ∈ q0) ⊦ ∃ n0. eq-entry(n0, lref) = eq-entry(m, lref0) ∨ eq-entry(n0, lref) ∈ q0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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