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Proof step 10 from proof copy2append0 (previous step)

# ar1 = (n2 + n1) + n0, n1 + n0 ≤ # ar1, n1 ≤ # ar0, n0 + n1 ≤ # ar1, n0 ≤ # ar1 ⊦ # copy(ar0, 0, ar1, n0, n1) = #(subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2)) ∧ (∀ n. n < # copy(ar0, 0, ar1, n0, n1) → copy(ar0, 0, ar1, n0, n1)[n] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 12:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 13:
# ar1 = (n2 + n1) + n0, n < # ar1, n1 + n0 ≤ # ar1, n1 ≤ # ar0, n0 + n1 ≤ # ar1, n0 ≤ # ar1 ⊦ (n < n0 ∨ n0 + n1 ≤ n ⊃ ar1[n];ar0[n - n0]) = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n]

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n + 0 = n; (s)%⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]); (s)%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)% ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s)