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Proof step 12 from proof valid-wl-erase (previous step)

wlar[n] = wl-entry(m4, stat0), apebs[n] = mkpeb(aehdr, avhdr, buf, m2, boolvar), wlar[m] = wl-entry(m0, stat), 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)) ⊦ n = m, stat = erroneous, (stat0 = erroneous ↔ boolvar) ∧ (stat0 = free → avhdr = empty) ∧ (stat0 ≠ erroneous ∧ stat0 ≠ erasing → aehdr = aechdr(m4)) ∧ (stat0 = used → (∃ v, m1. valid-peb(v, m1, mkpeb(aehdr, avhdr, buf, m2, false)) ∧ avhdr.sqn = valid-sqns(v, m1, apebs).max))

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).echdr = aehdr; (s)% ⊦ wl-entry(m, stat).ec = m; (s)% ⊦ used ≠ erroneous; (s)% ⊦ used ≠ free; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s) Used terms: n