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Proof step 38 from proof prepost-init-vols-mappings (previous step)

# wlar = # apebs, avhdr.leb = VTBL_LNUM, avhdr0.sqn = valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs).max, avhdr.vol = VTBL_VOLID, recs0[VTBL_VOLID × VTBL_LNUM] = recovery-entry(n0, m3), apebs[n] = mkpeb(aehdr, avhdr, buf, m, false), apebs[n0] = mkpeb(aechdr(m4), avhdr0, buf0, m1, false), wlar[n0] = wl-entry(m4, used), wlar[n] = wl-entry(m0, used), avhdr0.valid?, n0 < # apebs, n < # apebs, bflipset-cons(bflipset, wlar), is-encoded(to-vtbl(vols), buf), unique-sqns(apebs), unmapped-vols(vols), valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m4), avhdr0, buf0, m1, false)), VTBL_VOLID × VTBL_LNUM ∈ recs0, ∀ 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)) ⊦ n0 = n, VTBL_VOLID ∈ vols, n ∈ q

User applied rule all left

And obtained the following new goals

Proof step 39:
# wlar = # apebs, avhdr.leb = VTBL_LNUM, avhdr0.sqn = valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs).max, avhdr.vol = VTBL_VOLID, wlar[n] = wl-entry(m0, used), wlar[n0] = wl-entry(m4, used), apebs[n0] = mkpeb(aechdr(m4), avhdr0, buf0, m1, false), apebs[n] = mkpeb(aehdr, avhdr, buf, m, false), recs0[VTBL_VOLID × VTBL_LNUM] = recovery-entry(n0, m3), avhdr0.valid?, n < # apebs, n0 < # apebs, bflipset-cons(bflipset, wlar), is-encoded(to-vtbl(vols), buf), unique-sqns(apebs), unmapped-vols(vols), valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m4), avhdr0, buf0, m1, false)), VTBL_VOLID × VTBL_LNUM ∈ recs0, 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)) ⊦ n0 = n, n ∈ q, VTBL_VOLID ∈ vols

Rule argument

pos: left 21 terms: n abort: F discard: T computed: T

Additional information

Used terms: n