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Proof step 2 from proof append-at2 (previous step)

n < # ar0 + # ar1, # ar0 ≤ n ⊦ (ar0 + ar1)[n] = ar1[n - # ar0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # ar0 + # ar1, # ar0 ≤ n ⊦ # ar1 = 0, (ar0 + ar1)[n] = ar1[n - # ar0]

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic