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Proof step 13 from proof copy2append (previous step)

# ar1 = (n2 + m) + n1, n < # ar1, m + n1 ≤ # ar1, n0 + m ≤ # ar0, n1 + m ≤ # ar1, n1 ≤ # ar1 ⊦ (n < n1 ∨ n1 + m ≤ n ⊃ ar1[n];ar0[n + n0 - n1]) = (subarray(ar1, 0, n1) + subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]
Proof step 15:
# ar1 = (n2 + m) + n1, n < # ar1, n1 ≤ # ar1, n1 + m ≤ # ar1, n0 + m ≤ # ar0, m + n1 ≤ # ar1 ⊦ (n < n1 ∨ n1 + m ≤ n ⊃ ar1[n];ar0[n + n0 - n1]) = (subarray(ar1, 0, n1) + subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n], n < # subarray(ar1, 0, n1)

Rule argument

spec: <toplevel> name: append-at1 seq: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n] subst: [n, ar1, ar0] → [n, subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2), subarray(ar1, 0, n1)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s)% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n, subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2), subarray(ar1, 0, n1)