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Proof step 13 from proof mtd-cons-above-fillcount-empty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# 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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ n ≤ n0 → (m + (n0 - n)) + n = m + n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)