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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m1, n3, n0]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m1, n3, n0