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Proof step 23 from proof vols-is-max-is-min-eraseq (previous step)

# apebs = # wlar1, # wlar0 = # wlar1, wlar0[n] = wl-entry(m3, stat), apebs[n0] = mkpeb(aehdr, avhdr, buf, m2, boolvar), stat = erasing, n < # wlar0, bflipset-cons(bflipset0, wlar0), bflipset-cons(bflipset, wlar1), is-encoded(vtbl, buf), unique-sqns(apebs), valid-ebatbl(apebs, wlar0, vols0, n0), valid-ebatbl(apebs, wlar1, vols1, m0), valid-eraseq(apebs, q0), valid-eraseq(q0), valid-eraseq(apebs, q1), valid-eraseq(wlar1, q1), valid-eraseq(q1), valid-eraseq-strong(apebs, vtbl, q0), valid-global-sqn(apebs, n1), valid-global-sqn(apebs, m1), valid-vtbl(apebs, vols0, wlar0, n0), valid-vtbl(apebs, vols1, wlar1, m0), valid-wl(apebs, wlar0), valid-wl(apebs, wlar1), eq-entry(n, lref) ∈ q0, n ∈ q0, ∀ 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, VTBL_VOLID ∈ vols0, VTBL_VOLID ∈ vols1, lref ∈ q1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic