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Proof step 18 from proof subarray-split-first (previous step)

n0 + m < # ar, n < # mkarray(1)[0, ar[n0]] + # subarray(ar, n0 + 1, m), n ≤ m, # mkarray(1)[0, ar[n0]] ≤ n ⊦ ar[n0 + n] = subarray(ar, n0 + 1, m)[n - # mkarray(1)[0, ar[n0]]], n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n0 + m < # ar, 1 ≤ m, 1 ≤ n, n ≤ m ⊦ ar[n0 + n] = subarray(ar, n0 + 1, m)[n - 1], m = 0, n = 0

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (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 ↔ false); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # subarray(ar, m, n) = n; (s)% ⊦ # mkarray(n) = n; (s)% ⊦ # ar[n, a] = # ar; (s)