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Proof step 13 from proof copy2append0 (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 14:
# ar1 = (n2 + n1) + n0, n < # ar1, n0 ≤ # ar1, n0 + n1 ≤ # ar1, n1 ≤ # ar0, n1 + n0 ≤ # ar1, n < n0 ∨ n0 + n1 ≤ n ⊦ ar1[n] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n]
Proof step 34:
# ar1 = (n2 + n1) + n0, n < # ar1, n0 ≤ # ar1, n0 + n1 ≤ # ar1, n1 ≤ # ar0, n1 + n0 ≤ # ar1 ⊦ ar0[n - n0] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n], n < n0, n0 + n1 ≤ n

Rule argument

pos: right 1

Proof step 13 from proof copy2append0 (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information