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Proof step 5 from proof mtd-cons-above-fillcount-empty (previous step)

# buf = PEB_SIZE, pebs[n] = mkpeb(buf, m, false), n < # pebs, is-aligned(m, EB_PAGE_SIZE), range=(buf, m, emptypeb, m, PEB_SIZE - m), m0 + m1 ≤ PEB_SIZE, m ≤ m0, m ≤ PEB_SIZE ⊦ subarray(buf, m0, m1) = mkempbuf(m1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
# buf = PEB_SIZE, PEB_SIZE = n0 + m, pebs[n] = mkpeb(buf, m, false), n < # pebs, is-aligned(m, EB_PAGE_SIZE), range=(buf, m, emptypeb, m, n0), m ≤ PEB_SIZE, m ≤ m0, m0 + m1 ≤ PEB_SIZE ⊦ subarray(buf, m0, m1) = mkempbuf(m1)

Rule argument

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

Additional information

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