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Proof step 11 from proof copy-subarray2 (previous step)

n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ m1 = 0, copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2)[n] = (n < n1 ∨ n1 + n2 ≤ n ⊃ ar1[n + m0];ar0[n + n0 - n1]), # ar1 = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 12:
n < n1 ∨ n1 + n2 ≤ n, n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2)[n] = ar1[n + m0], m1 = 0, # ar1 = 0
Proof step 17:
¬(n < n1 ∨ n1 + n2 ≤ n), n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2)[n] = ar0[n + n0 - n1], m1 = 0, # ar1 = 0

Rule argument

pos: right 2

Proof step 11 from proof copy-subarray2 (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information