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Proof step 530 from proof prepost-scan-all (previous step)

# wlar = n0 + vtblpnum, # apebs = n0 + vtblpnum, apebs[vtblpnum] = mkpeb(aehdr0, avhdr0, buf0, m, false), nx < vtblpnum, bflipset-cons(bflipset, subarray(wlar, 0, vtblpnum)), indices-erasing(wlar, nx), ubi-io-inv(apebs), ubi-io-peb-inv(mkpeb(aehdr0, avhdr0, buf0, m, false)), unique-sqns(apebs), valid-eraseq(subarray(wlar, 0, vtblpnum), q), valid-eraseq-stronger(subarray(apebs, 0, vtblpnum), q), valid-global-sqn(subarray(apebs, 0, vtblpnum), sqnum), valid-recs(subarray(apebs, 0, vtblpnum), subarray(wlar, 0, vtblpnum), recs), valid-recs-sqns(apebs, recs), valid-wl(subarray(apebs, 0, vtblpnum), subarray(wlar, 0, vtblpnum)) ⊦ n0 = 0, dups(nx), ubi-io-read-echdr-pre(vtblpnum, apebs)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ ubi-io-read-echdr-pre(n, apebs) ↔ n < # apebs ∧ ¬ apebs[n].bad; (s)