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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ vols1. # vols1 < # vols0[v, ebatbl] → true → # to-vtbl(vols1) = # vols1 ⊦ # to-vtbl(vols0)[v, # ebatbl] = # vols0 + 1, v ∈ 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, ebatbl] = # vols + 1; (s)% ⊦ ¬ v ∈ vols → vols -- v = vols; (s)% ⊦ v ∈ vols[v0, ebatbl] ↔ v = v0 ∨ v ∈ vols; (ws)