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Proof step 12 from proof to-vtbl-at (previous step)

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], v = v0, v0 ∈ vols0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
v ∈ vols0, ∀ vols1, v. # vols1 < # vols0[v0, ebatbl] → v ∈ vols1 → to-vtbl(vols1)[v] = # vols1[v] ⊦ to-vtbl(vols0)[v] = # vols0[v], v0 ∈ vols0

Rule argument

Additional information

Used simplifier rules: ⊦ v ≠ v0 → (v ∈ vols[v0, ebatbl] ↔ v ∈ vols); (s)% ⊦ v ≠ v0 → vols[v0, ebatbl][v] = vols[v]; (s)% ⊦ v ≠ v0 → vtbl[v0, n][v] = vtbl[v]; (s)