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Proof step 13 from proof subarray-copy (previous step)

n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ m1 = 0, (n < n3 ∨ n3 + n2 ≤ n ⊃ ar1[n + m0];ar0[n + n0 - n3]) = copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n3, n2)[n], # ar1 = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 14:
n < n3 ∨ n3 + n2 ≤ n, n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ ar1[n + m0] = copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n3, n2)[n], m1 = 0, # ar1 = 0
Proof step 19:
¬(n < n3 ∨ n3 + n2 ≤ n), n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ ar0[n + n0 - n3] = copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n3, n2)[n], m1 = 0, # ar1 = 0

Rule argument

pos: right 2

Proof step 13 from proof subarray-copy (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information