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Proof step 6 from proof copy-append (previous step)

n0 < # copy(ar0, n, ar1, n1, m1), n0 < # ar1 ⊦ (n0 < n1 + m1 ∨ (n1 + m1) + m0 ≤ n0 ⊃ copy(ar0, n, ar1, n1, m1)[n0];ar0[n0 + n + m1 - n1 + m1]) = (n0 < n1 ∨ n1 + m0 + m1 ≤ n0 ⊃ ar1[n0];ar0[n0 + n - n1]), # ar1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n0 < # ar1 ⊦ # ar1 = 0, (n0 < n1 + m1 ∨ (n1 + m1) + m0 ≤ n0 ⊃ (n0 < n1 ∨ n1 + m1 ≤ n0 ⊃ ar1[n0];ar0[n0 + n - n1]);ar0[n0 + n - n1]) = (n0 < n1 ∨ n1 + m0 + m1 ≤ n0 ⊃ ar1[n0];ar0[n0 + n - n1])

Rule argument

Additional information

Used simplifier rules: at-copy: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]);%size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat