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Proof step 7 from proof valid-vtbl-sqns-notempty (previous step)

avhdr.leb = VTBL_LNUM, avhdr.vol = VTBL_VOLID, wlar[n0] = wl-entry(m1, used), apebs[n0] = mkpeb(aehdr, avhdr, buf, m0, boolvar), valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs) = ∅, n0 < # wlar, n0 < # apebs, is-encoded(to-vtbl(vols), buf), ∀ 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)) ⊦

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: n0

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ used ≠ erroneous; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.leb = m; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.vol = v; (s)% ⊦ used ≠ erasing; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).echdr = aehdr; (s)% ⊦ wl-entry(m, stat).ec = m; (s)% ⊦ used ≠ free; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%n < # apebs ⊦ apebs[n] = apeb ∧ valid-peb(v, m, apeb) → valid-sqns(v, m, apebs) ≠ ∅; (s) Used terms: n0