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Proof step 20 from proof range-copy-drop (previous step)

n0 + m0 ≤ # ar0, n + m0 ≤ # ar, n2 + m1 ≤ # ar2, n1 + m1 ≤ # ar0, ∀ n. n < m1 → ar2[n2 + n] = ar0[n1 + n], n0 + m0 ≤ n1 ∨ n1 + m1 ≤ n0 ⊦ n2 + m1 ≤ # ar2 ∧ n1 + m1 ≤ # copy(ar, n, ar0, n0, m0) ∧ (∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, ar0, n0, m0)[n1 + n3])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
n3 < m1, n0 + m0 ≤ # ar0, n + m0 ≤ # ar, n2 + m1 ≤ # ar2, n1 + m1 ≤ # ar0, ∀ n. n < m1 → ar2[n2 + n] = ar0[n1 + n], n0 + m0 ≤ n1 ∨ n1 + m1 ≤ n0 ⊦ ar2[n2 + n3] = copy(ar, n, ar0, n0, m0)[n1 + n3]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s)