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Proof step 22 from proof copy-shift-right (previous step)

n < # ar0 + # ar1, n < n0 + # ar0 ∨ (n0 + # ar0) + m1 ≤ n ⊦ (ar0 + ar1)[n] = copy(ar, m, ar1, n0, m1)[n - # ar0], n < # ar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0]
Proof step 24:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 25:
n < # ar0 + # ar1, # ar0 ≤ n, n < n0 + # ar0 ∨ (n0 + # ar0) + m1 ≤ n ⊦ # ar1 = 0, ar1[n - # ar0] = copy(ar, m, ar1, n0, m1)[n - # ar0]

Rule argument

Additional information

Used simplifier rules: append-at2: ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < m + n ↔ n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)