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Proof step 24 from proof copy2append (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
# ar1 = (n2 + m) + n1, n = n3 + n1, n < # ar1, n1 < # ar1, n1 ≤ n, n1 + m ≤ # ar1, n0 + m ≤ # ar0, m + n1 ≤ # ar1 ⊦ n2 + m = 0, (n1 + m ≤ n ⊃ ar1[n];ar0[n3 + n0]) = (subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n3]

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n1, n3]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n1, n3