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Proof step 15 from proof copy2append (previous step)

# ar1 = (n2 + m) + n1, n < # ar1, n1 ≤ # ar1, n1 + m ≤ # ar1, n0 + m ≤ # ar0, m + n1 ≤ # ar1 ⊦ (n < n1 ∨ n1 + m ≤ n ⊃ ar1[n];ar0[n + n0 - n1]) = (subarray(ar1, 0, n1) + subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n], n < # subarray(ar1, 0, n1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# ar1 = (n2 + m) + n1, n < # ar1, n1 < # ar1, m + n1 ≤ # ar1, n0 + m ≤ # ar0, n1 + m ≤ # ar1, n1 ≤ n ⊦ n2 + m = 0, (n1 + m ≤ n ⊃ ar1[n];ar0[(n - n1) + n0]) = (subarray(ar1, 0, n1) + subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (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% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # subarray(ar, m, n) = n; (s)