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Proof step 26 from proof copy2append (previous step)

# ar1 = (n2 + m) + n1, n3 < n2 + m, n1 < # ar1, m + n1 ≤ # ar1, n0 + m ≤ # ar0, n1 + m ≤ # ar1 ⊦ (m ≤ n3 ⊃ ar1[n3 + n1];ar0[n3 + n0]) = (subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n3]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]
Proof step 28:
# ar1 = (n2 + m) + n1, n1 < # ar1, n3 < n2 + m, n1 + m ≤ # ar1, n0 + m ≤ # ar0, m + n1 ≤ # ar1 ⊦ (m ≤ n3 ⊃ ar1[n3 + n1];ar0[n3 + n0]) = (subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n3], n3 < # subarray(ar0, n0, m)

Rule argument

spec: <toplevel> name: append-at1 seq: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n] subst: [n, ar1, ar0] → [n3, subarray(ar1, n1 + m, n2), subarray(ar0, n0, m)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s)%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n3, subarray(ar1, n1 + m, n2), subarray(ar0, n0, m)