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Proof step 19 from proof copy-shift-right (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;