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Proof step 18 from proof copy-subarray2 (previous step)

n1 < m1, n < n1 + n2, n1 ≤ n, n < m1, m0 + m1 ≤ # ar1, n + m0 < # ar1 ⊦ n1 + n2 = 0, copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2)[n] = ar0[(n - n1) + n0], # ar1 = 0, m1 = 0, n2 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
n = n3 + n1, n1 < m1, n < n1 + n2, n1 ≤ n, n < m1, m0 + m1 ≤ # ar1, n + m0 < # ar1 ⊦ n1 + n2 = 0, copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2)[n] = ar0[n3 + n0], # ar1 = 0, m1 = 0, n2 = 0

Rule argument

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

Additional information

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