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Proof step 28 from proof copy2append (previous step)

# ar1 = (n2 + m) + n1, n1 < # ar1, n3 < n2 + m, n1 + m ≤ # ar1, n0 + m ≤ # ar0, m + n1 ≤ # ar1 ⊦ (m ≤ n3 ⊃ ar1[n3 + n1];ar0[n3 + n0]) = (subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n3], n3 < # subarray(ar0, n0, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
# ar1 = (n2 + m) + n1, n1 < # ar1, n3 < n2 + m, m ≤ n3, n1 + m ≤ # ar1, n0 + m ≤ # ar0, m + n1 ≤ # ar1 ⊦ ar1[n3 + n1] = (subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n3], n2 = 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)