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Proof step 11 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, m + n0 ≤ # buf, m + n0 ≤ # emptypeb, ∀ n. n < n0 → buf[m + n] = (emptypeb)[m + n] ⊦ buf[m0 + n1] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# 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, ∀ n2. n2 < n0 → buf[m + n2] = empty ⊦ buf[m0 + n1] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%n < # emptypeb ⊦ (emptypeb)[n] = empty; (s)% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # emptypeb = PEB_SIZE; (s)