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Proof step 10 from proof in-rm-v-in (previous step)

lref0 = v0 × m0, eq-entry(n, lref) ∈ q0 - v0, ∀ n, lref, v. eq-entry(n, lref) ∈ q0 - v → (∃ n. eq-entry(n, lref) ∈ q0) ⊦ lref0 = none, ∃ n0. eq-entry(n0, lref) = eq-entry(m, lref0) ∨ eq-entry(n0, lref) ∈ q0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
eq-entry(n, lref) ∈ q0 - v0, ∀ n, lref, v. eq-entry(n, lref) ∈ q0 - v → (∃ n. eq-entry(n, lref) ∈ q0) ⊦ ∃ n0. eq-entry(n0, lref) = eq-entry(m, v0 × m0) ∨ eq-entry(n0, lref) ∈ q0

Rule argument

Additional information

Used simplifier rules: ⊦ none ≠ v × m; (s)