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Proof step 48 from proof prepost-scan-all (previous step)

# apebs = n0 + vtblpnum, # wlar = n0 + vtblpnum, apebs[vtblpnum] = mkpeb(aehdr, avhdr, buf, m, true), wlar[vtblpnum] = wl-entry(m0, stat), nx < vtblpnum, bflipset-cons(bflipset, subarray(wlar, 0, vtblpnum)), indices-erasing(wlar, nx), ubi-io-inv(apebs), ubi-io-peb-inv(mkpeb(aehdr, avhdr, buf, m, true)), 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), valid-global-sqn(subarray(apebs, 0, vtblpnum + 1), sqnum) ∧ valid-wl(subarray(apebs, 0, vtblpnum + 1), subarray(wlar[vtblpnum, wl-entry(m0, erroneous)], 0, vtblpnum + 1))

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%m < # apebs, # apebs = # wlar ⊦ valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)) → ( valid-wl (subarray(apebs, 0, m + 1), subarray(wlar[m, wl-entry(m0, erroneous)], 0, m + 1)) ↔ apebs[m].bad); (ss)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%m < # apebs ⊦ valid-global-sqn(subarray(apebs, 0, m), n) → ( valid-global-sqn(subarray(apebs, 0, m + 1), n) ↔ ¬ valid-peb(apebs[m]) ∨ apebs[m].vidhdr.sqn < n); (ss)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ valid-peb(mkpeb(aehdr, avhdr, buf, m, true)) ↔ false; (s)