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Proof step 3 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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
eq-entry(n, v × m) ∈ q ⊦ eq-entry(n, v × m) = eq-entry(n0, lref), v × m = lref, ∃ n. eq-entry(n, v × m) ∈ q

Rule argument

Additional information

Used simplifier rules: ⊦ eq-entry(m, lref) = eq-entry(m0, lref0) ↔ m = m0 ∧ lref = lref0; (ws)