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Proof step 15 from proof lf-mapped-not-bad (previous step)

vols[v0][m1] = ⌜n⌝ , apebs[n] = mkpeb(aehdr, avidhdr(v0, m1, m2, m3, m4), buf, m0, true), wlar[n] = wl-entry(m5, used), n < # apebs, n < # wlar, m1 < # vols[v0], v0 ∈ vols, ∀ v, m, n. vols[v][m] = ⌜n⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n].status = used ∧ n < # apebs ∧ n < # wlar ∧ apebs[n].vidhdr.valid? ∧ apebs[n].vidhdr.vol = v ∧ apebs[n].vidhdr.leb = m ∧ n ≠ 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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 9 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) Used terms: n