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Proof step 14 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n < # ar, n0 < # ar, n0 + n1 ≤ # ar, n0 ≤ n ⊦ n2 + n1 = 0, (n < n0 + n1 ⊃ a;ar[n]) = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n]

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, fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2), subarray(ar, 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)% ⊦ # fill(ar, a) = # ar; (s)% ⊦ # mkarray(n) = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ ¬ n < n; (s)%# ar = n ⊦ subarray(ar, 0, n) = ar; (s) Used terms: n, fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2), subarray(ar, 0, n0)