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Proof step 16 from proof subarray-split-first (previous step)

n0 + m < # ar, n ≤ m ⊦ ar[n0 + n] = (mkarray(1)[0, ar[n0]] + subarray(ar, n0 + 1, m))[n], n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0]
Proof step 18:
n0 + m < # ar, n < # mkarray(1)[0, ar[n0]] + # subarray(ar, n0 + 1, m), n ≤ m, # mkarray(1)[0, ar[n0]] ≤ n ⊦ ar[n0 + n] = subarray(ar, n0 + 1, m)[n - # mkarray(1)[0, ar[n0]]], n = 0

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(ar, n0 + 1, m), mkarray(1)[0, ar[n0]]] flags: <all paths>

Additional information

Used simplifier rules: ⊦ # ar[n, a] = # ar; (s)% ⊦ # mkarray(n) = n; (s)% ⊦ # subarray(ar, m, n) = n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s) Used terms: n, subarray(ar, n0 + 1, m), mkarray(1)[0, ar[n0]]