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Proof step 6 from proof to-vtbl-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
v0 ∈ vols, ∀ vols0, v. # vols0 < # vols → true → (v ∈ to-vtbl(vols0) ↔ v ∈ vols0) ⊦ v ∈ to-vtbl(vols) ↔ v ∈ vols

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ v ∈ vols → vols[v] = (∅)[v]; (s)% ⊦ ¬ v ∈ ∅; (s)