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Proof step 11 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n < # ar, n0 + n1 ≤ # ar ⊦ (n0 ≤ n ∧ 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

Proof step 12:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]
Proof step 13:
# ar = n2 + n0 + n1, n < # ar, n0 + n1 ≤ # ar ⊦ (n0 ≤ n ∧ n < n0 + n1 ⊃ a;ar[n]) = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n], n < # subarray(ar, 0, n0)

Rule argument

spec: <toplevel> name: append-at1 seq: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n] 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: ⊦ # subarray(ar, m, n) = n; (s)% ⊦ m < n → (n ≤ m ↔ false); (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, fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2), subarray(ar, 0, n0)