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Proof step 31 from proof valid-wl-next-free (previous step)

# apebs = # wlar, apebs[m] = mkpeb(aechdr(m0), empty, buf, m1, false), apebs[n] = mkpeb(aehdr, avhdr, buf0, m3, true), wlar[n] = wl-entry(m4, stat), n + 1 < # wlar, n < # wlar, m < # wlar, n < m, ∀ n. n < m → (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, m2. valid-peb(v, m2, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m2, apebs, m).max)) ⊦ n = m, stat = erroneous

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)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s) Used terms: n