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Proof step 14 from proof to-vtbl-in (previous step)

∀ vols1, v. # vols1 < # vols0[v0, ebatbl] → true → (v ∈ to-vtbl(vols1) ↔ v ∈ vols1) ⊦ v = v0 ∨ v ∈ to-vtbl(vols0) ↔ v = v0 ∨ v ∈ vols0, v0 ∈ vols0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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