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Proof step 15 from proof subarray-append-shift-right (previous step)

n < # ar0 + # ar1, n < n0 + # ar0, n0 ≤ # ar1 ⊦ (n < # ar0 ⊃ ar0[n];ar1[n - # ar0]) = (ar0 + subarray(ar1, 0, n0))[n], n < # ar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n < # ar0 + # ar1, n < n0 + # ar0, # ar0 ≤ n, n0 ≤ # ar1 ⊦ # ar1 = 0, ar1[n - # ar0] = (ar0 + subarray(ar1, 0, n0))[n], n0 = 0

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)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)