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Proof step 2 from proof valid-ebatbl-move-vtbl (previous step)

wlar[n1] = wl-entry(m1, free), valid-ebatbl(apebs, wlar, vols, n0) ⊦ valid-ebatbl(apebs[n1, apeb], wlar[n1, wle][n0, wl-entry(m0, erasing)], vols, n1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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 → apebs[n, apeb][m] = apebs[m]; (s)% ⊦ # wlar[n, wle] = # wlar; (s)% ⊦ # apebs[n, apeb] = # apebs; (s)