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Proof step 5 from proof lf-used-valid-peb (previous step)

wlar[n] = wl-entry(m, used), n < # 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)) ⊦ valid-peb(apebs[n])

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ used ≠ erroneous; (s)% ⊦ valid-peb(v, m, apeb) → valid-peb(apeb); (ss) Used terms: n