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Proof step 30 from proof vols-is-max-is-min-eraseq (previous step)

# wlar0 = # wlar1, # apebs = # wlar1, wlar1[n].status = erroneous ↔ boolvar0, apebs[n0] = mkpeb(aehdr, avhdr, buf, m2, boolvar), wlar0[n] = wl-entry(m3, erasing), apebs[n] = mkpeb(aehdr0, avhdr0, buf0, m4, boolvar0), n < # wlar1, bflipset-cons(bflipset, wlar1), bflipset-cons(bflipset0, wlar0), is-encoded(vtbl, buf), unique-sqns(apebs), valid-ebatbl(apebs, wlar1, vols1, m0), valid-ebatbl(apebs, wlar0, vols0, n0), valid-eraseq(q1), valid-eraseq(wlar1, q1), valid-eraseq(apebs, q1), valid-eraseq(q0), valid-eraseq(apebs, q0), valid-eraseq-strong(apebs, vtbl, q0), valid-global-sqn(apebs, m1), valid-global-sqn(apebs, n1), valid-vtbl(apebs, vols1, wlar1, m0), valid-vtbl(apebs, vols0, wlar0, n0), valid-wl(apebs, wlar0), n ∈ q0, eq-entry(n, lref) ∈ q0, wlar1[n].status = used → valid-peb(v, m, apebs[n]) ∧ avhdr0.sqn = valid-sqns(v, m, apebs).max, ¬ boolvar0 ∧ wlar1[n].status ≠ erasing → aehdr0 = aechdr(wlar1[n].ec), wlar1[n].status = free → avhdr0 = empty, ∀ v. (v ∈ vols1 ↔ v ∈ vols0) ∧ (v ∈ vols1 → # vols1[v] = # vols0[v] ∧ (∀ m. m < # vols1[v] ∧ vols1[v][m] ≠ unmapped → vols1[v][m] = vols0[v][m])) ⊦ lref = none, lref ∈ q1, VTBL_VOLID ∈ vols1, VTBL_VOLID ∈ vols0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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