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Proof step 2 from proof is-vtbl-pnum (previous step)

avhdr.leb = VTBL_LNUM, avhdr.vol = VTBL_VOLID, apebs[n] = mkpeb(aehdr, avhdr, buf, m, boolvar), wlar[n] = wl-entry(m0, used), n < # wlar, n < # apebs, valid-ebatbl(apebs, wlar, vols, n0), valid-wl(apebs, wlar) ⊦ n = n0, VTBL_VOLID ∈ vols

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
avhdr.leb = VTBL_LNUM, avhdr.vol = VTBL_VOLID, apebs[n] = mkpeb(aehdr, avhdr, buf, m, boolvar), wlar[n] = wl-entry(m0, used), n < # wlar, n < # apebs, ∀ 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, VTBL_VOLID ∈ vols

Rule argument

Additional information

Used simplifier rules: valid-wl: ⊦ valid-wl(apebs, wlar) ↔ (∀ 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)));%valid-ebatbl: ⊦ valid-ebatbl(apebs, wlar, vols, n0) ↔ (∀ 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);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic