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Proof step 7 from proof range-copy-inner (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
m1 + n1 = n4 + m0, n4 = n0 + n3 + n1, n3 + m0 ≤ m1, (n3 + n1) + m0 ≤ # ar0, n + m0 ≤ # ar, m0 ≤ m1 + n1 ⊦ range=(ar2, n2, copy(ar, n, ar0, n3 + n1, m0), n1, m1) ↔ range=(ar2, n2, ar0, n1, n3) ∧ range=(ar2, n3 + n2, ar, n, m0) ∧ range=(ar2, n3 + n2 + m0, ar0, (n3 + n1) + m0, n0)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m1 + n1, m0, n4]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ n + m ≤ m0 → ¬ n0 + m0 < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m1 + n1, m0, n4