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Proof step 3 from proof range-copy-lower (previous step)

n1 ≤ # ar0, n0 + m0 ≤ # ar0, n + m0 ≤ # ar, n0 + m0 ≤ n1 + m1, n1 ≤ n0 + m0, n0 ≤ n1, n0 ≤ # ar0 ⊦ range=(ar2, n2, copy(ar, n, ar0, n0, m0), n1, m1) ↔ range=(ar2, n2, ar, n + (n1 - n0), m0 + n0 - n1) ∧ range=(ar2, (n0 + m0 - n1) + n2, ar0, n0 + m0, m1 + n1 - m0 + n0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
n1 = n3 + n0, n0 ≤ # ar0, n0 ≤ n1, n1 ≤ n0 + m0, n0 + m0 ≤ n1 + m1, n + m0 ≤ # ar, n0 + m0 ≤ # ar0, n1 ≤ # ar0 ⊦ range=(ar2, n2, copy(ar, n, ar0, n0, m0), n1, m1) ↔ range=(ar2, n2, ar, n + n3, m0 + n0 - n1) ∧ range=(ar2, (n0 + m0 - n1) + n2, ar0, n0 + m0, m1 + n1 - m0 + n0)

Rule argument

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

Additional information

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