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Proof step 2 from proof valid-ebatbl-unmap (previous step)

vols[v][m] = ⌜n⌝ , m < # vols[v], valid-ebatbl(apebs, wlar, vols, n0), v ∈ vols ⊦ valid-ebatbl(apebs, wlar[n, wl-entry(m0, erasing)], vols[v, vols[v][m, unmapped]], n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
vols[v][m] = ⌜n⌝ , m < # vols[v], v ∈ vols, ∀ v, m, n1. vols[v][m] = ⌜n1⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n1].status = used ∧ n1 < # apebs ∧ n1 < # wlar ∧ apebs[n1].vidhdr.valid? ∧ apebs[n1].vidhdr.vol = v ∧ apebs[n1].vidhdr.leb = m ∧ n1 ≠ n0 ⊦ vols[v, vols[v][m, unmapped]][v0][m1] = ⌜n1⌝ ∧ v0 ∈ vols[v, vols[v][m, unmapped]] ∧ m1 < # vols[v, vols[v][m, unmapped]][v0] ↔ wlar[n, wl-entry(m0, erasing)][n1].status = used ∧ n1 < # apebs ∧ n1 < # wlar ∧ apebs[n1].vidhdr.valid? ∧ apebs[n1].vidhdr.vol = v0 ∧ apebs[n1].vidhdr.leb = m1 ∧ n1 ≠ n0

Rule argument

Additional information

Used simplifier rules: valid-ebatbl: ⊦ valid-ebatbl(apebs, wlar, vols, n0) ↔ (∀ v, m, n. vols[v][m] = ⌜n⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n].status = used ∧ n < # apebs ∧ n < # wlar ∧ apebs[n].vidhdr.valid? ∧ apebs[n].vidhdr.vol = v ∧ apebs[n].vidhdr.leb = m ∧ n ≠ n0);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # wlar[n, wle] = # wlar; (s)