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Proof step 20 from proof range-eq-splice-less (previous step)

ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m0), n0, m0)[n1 + n3], n3 < m0, n3 < m1, # ar0 < n0 + m1, # ar0 < n0 + m0, m1 ≤ m0, n + m0 ≤ # ar, n1 ≤ n0, n2 + m1 ≤ # ar2, n1 + m1 ≤ max(# ar0, n0 + m0) ⊦ copy(ar, n, resize(ar0, n0 + m0), n0, m0)[n1 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3], m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n ≤ n0 ∧ n1 ≤ m → n + n1 ≤ n0 + m; (s)% ⊦ n < m → max(n, m) = m; (s)% ⊦ m < n → n ≠ 0; (s)