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Proof step 6 from proof copy-subarray1-elim (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat