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Proof step 30 from proof inv-write (previous step)

PEB_SIZE = n1 + m2 + n, PEB_SIZE = n0 + m1, pebs[m3] = mkpeb(buf0, m1, false), m3 < # pebs, n3 < n1, is-aligned(m2, EB_PAGE_SIZE), is-aligned(n, EB_PAGE_SIZE), is-aligned(m1, EB_PAGE_SIZE), mtd-cons(pebs), m1 ≤ m2, n2 + n ≤ # buf, m2 + n ≤ PEB_SIZE, m1 ≤ PEB_SIZE, m1 + n0 ≤ # buf0, m1 + n0 ≤ # emptypeb, ∀ n. n < n0 → buf0[m1 + n] = (emptypeb)[m1 + n] ⊦ buf0[(m2 + n) + n3] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
PEB_SIZE = n1 + m2 + n, PEB_SIZE = n0 + m1, pebs[m3] = mkpeb(buf0, m1, false), m3 < # pebs, n3 < n1, is-aligned(m2, EB_PAGE_SIZE), is-aligned(n, EB_PAGE_SIZE), is-aligned(m1, EB_PAGE_SIZE), mtd-cons(pebs), m1 ≤ m2, n2 + n ≤ # buf, m2 + n ≤ PEB_SIZE, m1 ≤ PEB_SIZE, ∀ n4. n4 < n0 → buf0[m1 + n4] = empty ⊦ buf0[(m2 + n) + n3] = 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%n < # pebs ⊦ pebs[n] = mkpeb(buf, m, boolvar) ∧ mtd-cons(pebs) → # buf = PEB_SIZE; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # emptypeb = PEB_SIZE; (s)