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Proof step 9 from proof to-vtbl-in (previous step)

∃ ebatbl, vols0. vols = vols0[v0, ebatbl] ∧ ¬ v0 ∈ vols0, ∀ vols0, v. # vols0 < # vols → true → (v ∈ to-vtbl(vols0) ↔ v ∈ vols0) ⊦ v ∈ to-vtbl(vols) ↔ v ∈ vols

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ vols1, v. # vols1 < # vols0[v0, ebatbl] → true → (v ∈ to-vtbl(vols1) ↔ v ∈ vols1) ⊦ v ∈ to-vtbl(vols0)[v0, # ebatbl] ↔ v ∈ vols0[v0, ebatbl], v0 ∈ vols0

Rule argument

Additional information

Used simplifier rules: to-vtbl-rec: ⊦ to-vtbl(vols[v, ebatbl]) = to-vtbl(vols -- v)[v, # ebatbl];% ⊦ ¬ v ∈ vols → vols -- v = vols; (s)