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Proof step 20 from proof copy2append0 (previous step)

# ar1 = (n2 + n1) + n0, n < # ar1, n0 + n1 < # ar1, n0 + n1 ≤ n, n0 ≤ # ar1, n1 ≤ # ar0 ⊦ ar1[n] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n], n2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: append-at2 seq: ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0] subst: [n, ar1, ar0] → [n, subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2), subarray(ar1, 0, n0)] flags: <all paths>

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n];% ⊦ # subarray(ar, m, n) = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)%# ar = n ⊦ subarray(ar, 0, n) = ar; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat Used terms: n, subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2), subarray(ar1, 0, n0)