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Proof step 7 from proof is-vtbl-pnum (previous step)

avhdr.leb = VTBL_LNUM, avhdr.vol = VTBL_VOLID, apebs[n] = mkpeb(aehdr, avhdr, buf, m, boolvar), wlar[n] = wl-entry(m0, used), n < # wlar, n < # apebs, ∀ v, m, n1. vols[v][m] = ⌜n1⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n1].status = used ∧ n1 < # apebs ∧ n1 < # wlar ∧ apebs[n1].vidhdr.valid? ∧ apebs[n1].vidhdr.vol = v ∧ apebs[n1].vidhdr.leb = m ∧ n1 ≠ n0, ∀ 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 = n0, avhdr.valid?, VTBL_VOLID ∈ vols

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: n

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)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.valid?; (ss) Used terms: n