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Proof step 2 from proof in-prepend-vl (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∃ n. eq-entry(n, v × m) = eq-entry(n0, lref) ∨ eq-entry(n, v × m) ∈ q ⊦ v × m = lref, ∃ n. eq-entry(n, v × m) ∈ q
Proof step 5:
v × m = lref ∨ (∃ n. eq-entry(n, v × m) ∈ q) ⊦ ∃ n. eq-entry(n, v × m) = eq-entry(n0, lref) ∨ eq-entry(n, v × m) ∈ q

Rule argument

pos: right 1

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