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Proof step 3 from proof subarray-append-consec0 (conclusion)

⊦ n0 + n1 ≤ # ar → subarray(ar, 0, n0) + subarray(ar, n0, n1) = subarray(ar, 0, n0 + n1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n0 + n1 ≤ # ar ⊦ subarray(ar, 0, n0) + subarray(ar, n0, n1) = subarray(ar, 0, n0 + n1)

Rule argument

Additional information

[Concretesimprules(())]