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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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)

Rule argument

Additional information

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