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Proof step 9 from proof subarray-split-size (previous step)

⊦ # copy(ar, n, mkarray(n0 + n1), 0, n0 + n1) = #(copy(ar, n, mkarray(n0), 0, n0) + copy(ar, n + n0, mkarray(n1), 0, n1)) ∧ (∀ n2. n2 < # copy(ar, n, mkarray(n0 + n1), 0, n0 + n1) → copy(ar, n, mkarray(n0 + n1), 0, n0 + n1)[n2] = (copy(ar, n, mkarray(n0), 0, n0) + copy(ar, n + n0, mkarray(n1), 0, n1))[n2])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 11:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 12:
n2 < n0 + n1 ⊦ copy(ar, n, mkarray(n0 + n1), 0, n0 + n1)[n2] = (copy(ar, n, mkarray(n0), 0, n0) + copy(ar, n + n0, mkarray(n1), 0, n1))[n2]

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)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # mkarray(n) = n; (s)% ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s)