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Proof step 5 from proof copy-subarray1-elim (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
¬(n < m0 ∨ m0 + m1 ≤ n), n0 + n1 ≤ # ar0, m + m1 ≤ n1, n < # ar1 ⊦ copy(ar0, n0, mkarray(n1), 0, n1)[n + m - m0] = (n < m0 ∨ m0 + m1 ≤ n ⊃ ar1[n];ar0[n + n0 + m - m0]), # ar1 = 0

Rule argument

pos: right 1

Proof step 5 from proof copy-subarray1-elim (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information