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Proof step 30 from proof copy2append0 (previous step)

# ar1 = (n2 + n1) + n0, n3 < n2 + n1, n0 + n1 < # ar1, n0 < # ar1, n1 ≤ # ar0, n1 ≤ n3 ⊦ ar1[n3 + n0] = (subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n3], n2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 31:
⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0]
Proof step 32:
# ar1 = (n2 + n1) + n0, n0 < # ar1, n0 + n1 < # ar1, n3 < n2 + n1, n3 < # subarray(ar0, 0, n1) + # subarray(ar1, n0 + n1, n2), n1 ≤ n3, n1 ≤ # ar0, # subarray(ar0, 0, n1) ≤ n3 ⊦ ar1[n3 + n0] = subarray(ar1, n0 + n1, n2)[n3 - # subarray(ar0, 0, n1)], n2 = 0

Rule argument

spec: <toplevel> name: append-at2 seq: ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0] subst: [n, ar1, ar0] → [n3, subarray(ar1, n0 + n1, n2), subarray(ar0, 0, n1)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n3, subarray(ar1, n0 + n1, n2), subarray(ar0, 0, n1)