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Proof step 1 from proof subarray-subarray-elim (conclusion)

⊦ n0 + n1 ≤ # ar0 ∧ m0 + m1 ≤ n1 → subarray(subarray(ar0, n0, n1), m0, m1) = subarray(ar0, n0 + m0, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n0 + n1 ≤ # ar0, m0 + m1 ≤ n1 ⊦ copy(copy(ar0, n0, mkarray(n1), 0, n1), m0, mkarray(m1), 0, m1) = copy(ar0, n0 + m0, mkarray(m1), 0, m1)

Rule argument

Additional information

Used simplifier rules: Subarray: ⊦ subarray(ar, m, n) = copy(ar, m, mkarray(n), 0, n);