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Proof step 17 from proof lf-valid-eraseq-stronger-notin-recs (previous step)

# apebs = # wlar, avhdr.sqn = valid-sqns(v0, m1, apebs).max, apebs[n] = mkpeb(aechdr(m2), avhdr, buf, m0, false), wlar[n] = wl-entry(m2, used), n < # wlar, valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(v0, m1, mkpeb(aechdr(m2), avhdr, buf, m0, false)), ∀ ladr, m. ladr ∈ recs ∧ recs[ladr].pnum = m ↔ wlar[m].status = used ∧ m < # apebs ∧ apebs[m].vidhdr.vol = ladr.vol ∧ apebs[m].vidhdr.leb = ladr.leb ⊦ avhdr = empty, v0 × m1 ∈ recs, n ∈ q

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# 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)), ∀ ladr, m. ladr ∈ recs ∧ recs[ladr].pnum = m ↔ wlar[m].status = used ∧ m < # wlar ∧ apebs[m].vidhdr.vol = ladr.vol ∧ apebs[m].vidhdr.leb = ladr.leb ⊦ avhdr = empty, n ∈ q, v0 × m1 ∈ recs

Proof step 17 from proof lf-valid-eraseq-stronger-notin-recs (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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