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Proof step 10 from proof copy-subarray1-elim (previous step)

n < m0 + m1, m0 < # ar1, n < # ar1, m0 ≤ n, n0 + n1 ≤ # ar0, m + m1 ≤ n1 ⊦ copy(ar0, n0, mkarray(n1), 0, n1)[(n - m0) + m] = ar0[(n - m0) + n0 + m], m1 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n = n2 + m0, n < # ar1, m0 < # ar1, n < m0 + m1, m + m1 ≤ n1, n0 + n1 ≤ # ar0, m0 ≤ n ⊦ copy(ar0, n0, mkarray(n1), 0, n1)[n2 + m] = ar0[n2 + n0 + m], m1 = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [m0, n, n2]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, n, n2