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Proof step 5 from proof copy-coalesce (previous step)

m0 ≤ # ar1, m0 + n ≤ # ar1, n1 < # ar2 ⊦ (n1 < n2 ∨ n2 + m0 ≤ n1 ⊃ ar2[n1];copy(ar0, n0, ar1, 0, # ar1)[n1 + n - n2]) = (n1 < n2 ∨ n2 + m0 ≤ n1 ⊃ ar2[n1];ar0[n1 + n0 + n - n2]), # ar2 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 6:
¬(n1 < n2 ∨ n2 + m0 ≤ n1), m0 ≤ # ar1, m0 + n ≤ # ar1, n1 < # ar2 ⊦ copy(ar0, n0, ar1, 0, # ar1)[n1 + n - n2] = (n1 < n2 ∨ n2 + m0 ≤ n1 ⊃ ar2[n1];ar0[n1 + n0 + n - n2]), # ar2 = 0

Rule argument

pos: right 1

Proof step 5 from proof copy-coalesce (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information