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Proof step 10 from proof at-copy (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n0 < # ar1, m0 < n0 + m1, n0 ≤ m0, m0 < # ar1, ∀ m0, ar1, ar0, n, n0. m0 < # ar1 → copy(ar0, n, ar1, n0, m1)[m0] = (m0 < n0 ∨ n0 + m1 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]) ∨ # ar1 = 0 ⊦ n0 + m1 = 0, copy(ar0, n, ar1, n0, m1)[m0] = ar0[(m0 - n0) + n], # ar1 = 0, m0 = n0 + m1, 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%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1%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