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Proof step 13 from proof valid-wl-ebatbl-used-imp-mapped (previous step)

# wlar = # apebs, avhdr.sqn = valid-sqns(v, m, apebs).max, vols[v][# apebs] = ⌜n⌝ ∧ v ∈ vols ∧ # apebs < # vols[v] ↔ avhdr.valid? ∧ m = # apebs, wlar[n] = wl-entry(m1, used), apebs[n] = mkpeb(aechdr(m1), avhdr, buf, m0, false), n < # apebs, valid-peb(v, m, mkpeb(aechdr(m1), avhdr, buf, m0, false)), ∀ n. n < # apebs → (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)), ∀ v, m, n1. vols[v][m] = ⌜n1⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n1].status = used ∧ n1 < # apebs ∧ apebs[n1].vidhdr.valid? ∧ apebs[n1].vidhdr.vol = v ∧ apebs[n1].vidhdr.leb = m ∧ n1 ≠ n0 ⊦ n0 = n, v ∈ vols ∧ m < # vols[v] ∧ vols[v][m] = ⌜n⌝

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: v, m, n

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (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.leb = m; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.valid?; (ss) Used terms: v, m, n