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Proof step 6 from proof in-prepend-vl (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 7:
v × m = lref ⊦ ∃ n. eq-entry(n, v × m) = eq-entry(n0, lref) ∨ eq-entry(n, v × m) ∈ q
Proof step 9:
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: left 1

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