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Proof step 12 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, 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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
# apebs = # wlar, avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, false), wlar[n] = wl-entry(m2, stat), 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 = used → v = v0 ∧ m = m1 ∧ avhdr.sqn = valid-sqns(v0, m1, apebs).max, stat ≠ erasing → aehdr = aechdr(m2) ⊦ avhdr = empty, stat = erroneous, stat = free, v × m ∈ recs

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[n], m2, stat]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[n], m2, stat