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Proof step 2 from proof valid-wl-erasing-invalidate (previous step)

# apebs = # wlar, wlar[m] = wl-entry(m0, erasing), m < # apebs, valid-wl(apebs, wlar) ⊦ apeb.bad, valid-peb(apeb), valid-wl(apebs[m, apeb], wlar)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# apebs = # wlar, wlar[m] = wl-entry(m0, erasing), n < # wlar, m < # wlar, ∀ n. n < # wlar → (wlar[n].status = erroneous ↔ apebs[n].bad) ∧ (wlar[n].status = free → apebs[n].vidhdr = empty) ∧ (wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[n].echdr = aechdr(wlar[n].ec)) ∧ (wlar[n].status = used → (∃ v, m. valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max)) ⊦ apeb.bad, valid-peb(apeb), (wlar[n].status = erroneous ↔ apebs[m, apeb][n].bad) ∧ (wlar[n].status = free → apebs[m, apeb][n].vidhdr = empty) ∧ (wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[m, apeb][n].echdr = aechdr(wlar[n].ec)) ∧ (wlar[n].status = used → (∃ v, m1. valid-peb(v, m1, apebs[m, apeb][n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m1, apebs[m, apeb]).max))

Rule argument

Additional information

Used simplifier rules: valid-wl: ⊦ valid-wl(apebs, wlar) ↔ (∀ n. n < # wlar → (wlar[n].status = erroneous ↔ apebs[n].bad) ∧ (wlar[n].status = free → apebs[n].vidhdr = empty) ∧ ( wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[n].echdr = aechdr(wlar[n].ec)) ∧ ( wlar[n].status = used → (∃ v, m. valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max)));% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%m ≠ n ⊦ apebs[m, apeb][n] = apebs[n]; (s)% ⊦ valid-peb(v, m, apeb) → valid-peb(apeb); (ss)% ⊦ m < # apebs → apebs[m, apeb][m] = apeb; (s)