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Proof step 21 from proof append-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0]
Proof step 23:
n < (# ar + # ar0) + # ar1, n < # ar + # ar0, # ar ≤ n ⊦ # ar0 = 0, # ar0 + # ar1 = 0, ar0[n - # ar] = (ar + ar0 + ar1)[n]

Rule argument

Additional information

Used simplifier rules: append-at2: ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0];% ⊦ 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)