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Proof step 9 from proof mapped-pebs-add-empty-vol (previous step)

vols[v, fill(mkebatbl(n), unmapped)][v0][m] = ⌜n0⌝ , m < # vols[v, fill(mkebatbl(n), unmapped)][v0], v0 ∈ vols[v, fill(mkebatbl(n), unmapped)] ⊦ v0 = v, v ∈ vols, ∃ v, m. v ∈ vols ∧ m < # vols[v] ∧ vols[v][m] = ⌜n0⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
vols[v0][m] = ⌜n0⌝ , m < # vols[v0], v0 ∈ vols ⊦ v ∈ vols, ∃ v, m. v ∈ vols ∧ m < # vols[v] ∧ vols[v][m] = ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ v ∈ vols ∧ ¬ v0 ∈ vols → vols[v0, ebatbl][v] = vols[v]; (s)% ⊦ v ≠ v0 → (v ∈ vols[v0, ebatbl] ↔ v ∈ vols); (s)% ⊦ v ≠ v0 → vols[v0, ebatbl][v] = vols[v]; (s)