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Proof step 8 from proof copy-coalesce (previous step)

n1 = n3 + n2, n2 < # ar2, n1 < n2 + m0, n2 ≤ n1, n1 < # ar2, m0 + n ≤ # ar1, m0 ≤ # ar1 ⊦ n2 + m0 = 0, copy(ar0, n0, ar1, 0, # ar1)[n3 + n] = ar0[n3 + n0 + n], # ar2 = 0, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n3 < # ar1, n3 < m0, n2 < # ar2, m0 ≤ # ar1, m0 + n ≤ # ar1, n3 + n2 < # ar2 ⊦ m0 = 0, # ar2 = 0, copy(ar0, n0, ar1, 0, # ar1)[n3 + n] = ar0[n3 + n0 + n], n2 + m0 = 0, # ar1 = 0

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lep: ⊦ m ≤ m + n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2