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Proof step 7 from proof in-vl-append0 (previous step)

⊦ (∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, none) ∈ q), none = lref0

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 8:
∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0) ⊦ none = lref0, ∃ n. eq-entry(n, none) ∈ q
Proof step 10:
∃ n. eq-entry(n, none) ∈ q ⊦ none = lref0, ∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0)

Rule argument

pos: right 1

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