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Proof step 5 from proof range-copy-lower (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
m0 = n1 + n3, n0 ≤ # ar0, n3 + n0 ≤ # ar0, n0 + m0 ≤ # ar0, n + m0 ≤ # ar, m0 ≤ n3 + m1, n3 ≤ m0 ⊦ range=(ar2, n2, copy(ar, n, ar0, n0, m0), n3 + n0, m1) ↔ range=(ar2, n2, ar, n + n3, n1) ∧ range=(ar2, n1 + n2, ar0, n0 + m0, m1 + n3 - m0)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, n3, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, n3, n1