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Proof step 14 from proof lf-valid-eraseq-stronger-notin-recs (previous step)

# apebs = # wlar, avhdr.sqn = valid-sqns(v, m, apebs).max, wlar[n] = wl-entry(m2, stat), apebs[n] = mkpeb(aehdr, avhdr, buf, m0, false), stat = used, 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), stat ≠ erasing → aehdr = aechdr(m2), stat = used → v = v0 ∧ m = m1 ∧ avhdr.sqn = valid-sqns(v0, m1, apebs).max ⊦ avhdr = empty, stat = free, stat = erroneous, v × m ∈ recs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# apebs = # wlar, avhdr.sqn = valid-sqns(v0, m1, apebs).max, wlar[n] = wl-entry(m2, used), apebs[n] = mkpeb(aechdr(m2), avhdr, buf, m0, false), n < # wlar, valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(v0, m1, mkpeb(aechdr(m2), avhdr, buf, m0, false)), valid-recs(apebs, wlar, recs) ⊦ avhdr = empty, n ∈ q, v0 × m1 ∈ recs

Rule argument

Additional information

Used simplifier rules: lf-used-info: ⊦ n < # wlar ∧ wlar[n] = wl-entry(m, used) ∧ valid-eraseq(wlar, q) → ¬ n ∈ q;% ⊦ used ≠ erasing; (s)% ⊦ used ≠ free; (s)% ⊦ used ≠ erroneous; (s)