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Proof step 2 from proof range-copy-lower (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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)

Rule argument

Additional information

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