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Proof step 3 from proof in-vl-append (previous step)

⊦ (∃ n0. eq-entry(n0, lref) ∈ q ∨ eq-entry(n0, lref) = eq-entry(n, lref0)) ↔ lref = lref0, ∃ n. eq-entry(n, lref) ∈ q

User applied rule case distinction

And obtained the following new goals

Proof step 4:
∃ n0. eq-entry(n0, lref) ∈ q ∨ eq-entry(n0, lref) = eq-entry(n, lref0) ⊦ lref = lref0, ∃ n. eq-entry(n, lref) ∈ q
Proof step 6:
lref = lref0 ⊦ ∃ n. eq-entry(n, lref) ∈ q, ∃ n0. eq-entry(n0, lref) ∈ q ∨ eq-entry(n0, lref) = eq-entry(n, lref0)

Rule argument

pos: right 1

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