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Proof step 16 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, (n2 + m) + m1 ≤ PEB_SIZE, n0 ≤ n2 + n1 ⊦ buf[(n2 + m) + n1] = empty

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat name: anti seq: ⊦ m ≤ n → (n ≤ m ↔ m = n) subst: [m, n] → [PEB_SIZE, (n2 + m) + m1] flags: <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → ¬ n < m; (s) Used terms: PEB_SIZE, (n2 + m) + m1