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Proof step 1 from proof copy-subarray2 (conclusion)

# ar1 ≥ m0 + m1 ⊦ copy(ar0, n0, subarray(ar1, m0, m1), n1, n2) = subarray(copy(ar0, n0, ar1, m0 + n1, n2), m0, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m0 + m1 ≤ # ar1 ⊦ copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2) = copy(copy(ar0, n0, ar1, m0 + n1, n2), m0, mkarray(m1), 0, m1)

Rule argument

Additional information

Used simplifier rules: Subarray: ⊦ subarray(ar, m, n) = copy(ar, m, mkarray(n), 0, n);%ge: ⊦ n ≥ m ↔ m ≤ n; from specification nat