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Proof step 10 from proof to-vtbl-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
v ∈ vols0[v0, ebatbl], ∀ vols1, v. # vols1 < # vols0[v0, ebatbl] → v ∈ vols1 → to-vtbl(vols1)[v] = # vols1[v] ⊦ to-vtbl(vols0)[v0, # ebatbl][v] = # vols0[v0, ebatbl][v], 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)