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Proof step 10 from proof valid-wl-erase (previous step)

wlar[m] = wl-entry(m0, stat), apebs[m] = mkpeb(aehdr, avhdr, buf, m2, true), 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)) ⊦ stat = erroneous

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s) Used terms: m