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Proof step 5 from proof to-vtbl-update-vol (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ 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)