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Proof step 11 from proof lf-valid-eraseq-stronger-notin-recs (previous step)

# apebs = # wlar, avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # wlar, valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-recs(apebs, wlar, recs), 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 ∈ recs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# apebs = # wlar, avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, false), n < # wlar, valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, false)), valid-recs(apebs, wlar, recs), wlar[n].status ≠ erasing → aehdr = aechdr(wlar[n].ec), wlar[n].status = used → v = v0 ∧ m = m1 ∧ avhdr.sqn = valid-sqns(v0, m1, apebs).max ⊦ wlar[n].status = free, wlar[n].status = erroneous, avhdr = empty, v × m ∈ recs

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ valid-peb(v, m, mkpeb(aehdr, empty, buf, m0, boolvar)); (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).echdr = aehdr; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ ¬ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, true)); (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)