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Proof step 15 from proof lf-valid-recs-max (previous step)

# apebs = # wlar, wlar[m0] = wl-entry(m2, used), recs[v × n] = recovery-entry(m0, avhdr.sqn), apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # wlar, m0 < m, valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m1, boolvar)), valid-recs-sqns(apebs, recs), v × n ∈ recs, m ≤ # wlar, ∀ n0. n0 < m → (wlar[n0].status = erroneous ↔ apebs[n0].bad) ∧ (wlar[n0].status = free → apebs[n0].vidhdr = empty) ∧ (wlar[n0].status ≠ erroneous ∧ wlar[n0].status ≠ erasing → apebs[n0].echdr = aechdr(wlar[n0].ec)) ∧ (wlar[n0].status = used → (∃ v0, m3. valid-peb(v0, m3, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = valid-sqns(v0, m3, apebs, m).max)) ⊦ avhdr.sqn = valid-sqns(v, n, apebs, m).max

User applied rule all left

And closed the goal!

Rule argument

pos: left 11 terms: m0 abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ used ≠ erroneous; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s) Used terms: m0