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Proof step 35 from proof copy2append0 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 36:
# ar1 = (n2 + n1) + n0, n = n3 + n0, n0 < # ar1, n < # ar1, n < n0 + n1, n0 ≤ n, n0 + n1 ≤ # ar1, n1 ≤ # ar0, n1 + n0 ≤ # ar1 ⊦ n2 + n1 = 0, ar0[n3] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n], n1 = 0

Rule argument

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

Additional information

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