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Proof step 13 from proof subarray-append-consec (previous step)

n < n0 + n1, n < # subarray(ar, m0, n0) + # subarray(ar, n0 + m0, n1), m0 + n0 + n1 ≤ # ar, n0 ≤ n, # subarray(ar, m0, n0) ≤ n ⊦ subarray(ar, n0 + m0, n1)[n - # subarray(ar, m0, n0)] = ar[m0 + n], n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n < n0 + n1, n0 ≤ n, m0 + n0 + n1 ≤ # ar ⊦ subarray(ar, n0 + m0, n1)[n - n0] = ar[m0 + n], n1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s)