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Proof step 6 from proof range-copy-inner (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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)

Rule argument

Additional information

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