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Proof step 4 from proof valid-wl-ebatbl-used-imp-valid-ehdr (previous step)

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, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), vols[v][m] = ⌜n⌝ , m < # vols[v], v ∈ 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 ⊦ aehdr.valid?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
wlar[n].status = used, vols[avhdr.vol][avhdr.leb] = ⌜n⌝ , apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), avhdr.valid?, n < # apebs, n < # wlar, avhdr.leb < # vols[avhdr.vol], avhdr.vol ∈ 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

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic