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Proof step 2 from proof valid-ebatbl-update-wlar-unused (previous step)

wlar[m] = wl-entry(m0, stat) ⊦ valid-ebatbl(apebs, wlar[m, wl-entry(m1, stat0)], vols, n) ↔ valid-ebatbl(apebs, wlar, vols, n), stat = used, stat0 = used

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
wlar[m] = wl-entry(m0, stat) ⊦ (∀ v, m2, n0. vols[v][m2] = ⌜n0⌝ ∧ v ∈ vols ∧ m2 < # vols[v] ↔ wlar[m, wl-entry(m1, stat0)][n0].status = used ∧ n0 < # apebs ∧ n0 < # wlar ∧ apebs[n0].vidhdr.valid? ∧ apebs[n0].vidhdr.vol = v ∧ apebs[n0].vidhdr.leb = m2 ∧ n0 ≠ n) ↔ (∀ 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 ≠ n), stat = used, stat0 = used

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);% ⊦ # wlar[n, wle] = # wlar; (s)