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Proof step 35 from proof range-copy-upper (previous step)

n0 ≤ m0, n + m0 ≤ # ar, (n3 + n1) + m0 ≤ # ar0, n1 + n0 + n3 ≤ # ar0, n1 ≤ # ar0, n3 + n1 ≤ # ar0, n2 + n3 ≤ # ar2, n1 + n3 ≤ # ar0, (n3 + n2) + n0 ≤ # ar2, n + n0 ≤ # ar, ∀ n. n < n3 → ar2[n2 + n] = ar0[n1 + n], ∀ n4. n4 < n0 → ar2[(n3 + n2) + n4] = ar[n + n4] ⊦ n2 + n0 + n3 ≤ # ar2 ∧ (n4 < n0 + n3 → ar2[n2 + n4] = copy(ar, n, ar0, n3 + n1, m0)[n1 + n4])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
n4 < n0 + n3, n0 ≤ m0, n + m0 ≤ # ar, (n3 + n1) + m0 ≤ # ar0, n1 + n0 + n3 ≤ # ar0, n1 ≤ # ar0, n3 + n1 ≤ # ar0, n2 + n3 ≤ # ar2, n1 + n3 ≤ # ar0, (n3 + n2) + n0 ≤ # ar2, n + n0 ≤ # ar, ∀ n. n < n3 → ar2[n2 + n] = ar0[n1 + n], ∀ n4. n4 < n0 → ar2[(n3 + n2) + n4] = ar[n + n4] ⊦ ar2[n2 + n4] = copy(ar, n, ar0, n3 + n1, m0)[n1 + n4]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%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