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Proof step 14 from proof mtd-cons-above-fillcount-empty (previous step)

# buf = PEB_SIZE, PEB_SIZE = n0 + m, pebs[n] = mkpeb(buf, m, false), n < # pebs, n1 < m1, is-aligned(m, EB_PAGE_SIZE), m ≤ PEB_SIZE, m ≤ m0, m0 + m1 ≤ PEB_SIZE, n0 ≤ (m0 - m) + n1 ⊦ buf[m0 + n1] = empty

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
# buf = PEB_SIZE, PEB_SIZE = n0 + m, pebs[n] = mkpeb(buf, m, false), m0 = n2 + m, n1 < m1, n < # pebs, is-aligned(m, EB_PAGE_SIZE), n0 ≤ n2 + n1, m0 + m1 ≤ PEB_SIZE, m ≤ m0, m ≤ PEB_SIZE ⊦ buf[m0 + n1] = empty

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, m, n2]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, m, n2