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Proof step 4 from proof copy-coalesce (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 ≤ # ar1, m0 + n ≤ # ar1, n1 < # ar2 ⊦ (n1 < n2 ∨ n2 + m0 ≤ n1 ⊃ ar2[n1];copy(ar0, n0, ar1, 0, # ar1)[n1 + n - n2]) = (n1 < n2 ∨ n2 + m0 ≤ n1 ⊃ ar2[n1];ar0[n1 + n0 + n - n2]), # ar2 = 0

Rule argument

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%at-copy: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1