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Proof step 9 from proof valid-wl-ebatbl-used-imp-valid-ehdr (previous step)

wlar[n] = wl-entry(m5, used), apebs[n] = mkpeb(aehdr, avidhdr(v0, m1, m2, m3, m4), buf, m0, boolvar), vols[v0][m1] = ⌜n⌝ , m1 < # vols[v0], n < # wlar, n < # apebs, v0 ∈ vols, ∀ 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)), ∀ 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 = n0, aehdr.valid?

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 10:
vols[v0][m1] = ⌜n⌝ ∧ v0 ∈ vols ∧ m1 < # vols[v0] ↔ wlar[n].status = used ∧ n < # apebs ∧ n < # wlar ∧ apebs[n].vidhdr.valid? ∧ apebs[n].vidhdr.vol = v0 ∧ apebs[n].vidhdr.leb = m1 ∧ n ≠ n0, vols[v0][m1] = ⌜n⌝ , apebs[n] = mkpeb(aehdr, avidhdr(v0, m1, m2, m3, m4), buf, m0, boolvar), wlar[n] = wl-entry(m5, used), n < # apebs, n < # wlar, m1 < # vols[v0], v0 ∈ vols, ∀ 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, aehdr.valid?

Rule argument

pos: left 9 terms: v0, m1, n

Additional information

Used terms: v0, m1, n