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Proof step 14 from proof subarray-split-at (previous step)

# ar = n0 + n, n1 < n0 + n ⊦ ar[n1] = (subarray(ar, 0, n) + subarray(ar, n, n0))[n1], n0 = 0, n1 < # subarray(ar, 0, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# ar = n0 + n, n1 < n0 + n, n ≤ n1 ⊦ ar[n1] = (subarray(ar, 0, n) + subarray(ar, n, n0))[n1], 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)% ⊦ # subarray(ar, m, n) = n; (s)