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Proof step 4 from proof range-copy-lower (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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)

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m ≤ m + n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n - n + n1 = m - n1; (s)