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Proof step 25 from proof copy2append0 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 27:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 28:
# ar1 = (n2 + n1) + n0, n0 + n1 < # ar1, n < # ar1, n0 < # ar1, n0 ≤ n, n0 + n1 ≤ n, n1 ≤ # ar0 ⊦ n2 + n1 = 0, ar1[n] = (subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n - n0], n2 = 0

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ 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% ⊦ # subarray(ar, m, n) = n; (s)