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Proof step 45 from proof range-copy-lower (previous step)

n5 < n4 + n1, (n3 + n0) + n5 < # ar0, (n0 + n1 + n3) + n4 ≤ # ar0, (n1 + n2) + n4 ≤ # ar2, (n + n3) + n1 ≤ # ar, n2 + n1 ≤ # ar2, n + n1 + n3 ≤ # ar, n0 + n1 + n3 ≤ # ar0, n3 + n0 ≤ # ar0, n0 ≤ # ar0, ∀ n. n < n4 → ar2[(n1 + n2) + n] = ar0[(n0 + n1 + n3) + n], ∀ n4. n4 < n1 → ar2[n2 + n4] = ar[(n + n3) + n4], ∃ n0. n1 + n0 = n5 ⊦ ar2[n2 + n5] = ar0[(n3 + n0) + n5], n4 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
n6 < n4, (n3 + n0) + n1 + n6 < # ar0, (n0 + n1 + n3) + n4 ≤ # ar0, (n1 + n2) + n4 ≤ # ar2, (n + n3) + n1 ≤ # ar, n2 + n1 ≤ # ar2, n + n1 + n3 ≤ # ar, n0 + n1 + n3 ≤ # ar0, n3 + n0 ≤ # ar0, n0 ≤ # ar0, ∀ n. n < n4 → ar2[(n1 + n2) + n] = ar0[(n0 + n1 + n3) + n], ∀ n4. n4 < n1 → ar2[n2 + n4] = ar[(n + n3) + n4] ⊦ ar2[n2 + n1 + n6] = ar0[(n3 + n0) + n1 + n6]

Rule argument

Additional information

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