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Proof step 7 from proof to-vtbl-update (previous step)

v ∈ vols ⊦ v0 = v, (v0 ∈ to-vtbl(vols -- v)[v, # vols[v]] ↔ v0 ∈ vols) ∧ to-vtbl(vols -- v)[v, # vols[v]][v0] = to-vtbl(vols)[v0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
v ∈ vols ⊦ to-vtbl(vols -- v)[v0] = to-vtbl(vols)[v0], v0 = v

Rule argument

Additional information

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