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Proof step 10 from proof copy-other-weak (previous step)

∀ n1, ar1, n0, m0, m, ar, ar0. n1 < # ar1 ∧ n1 < n0 ∧ m0 + n0 ≤ m ∧ m + n2 ≤ # ar ∧ n0 ≤ # ar1 ∧ n2 ≤ # ar0 ∧ # ar ≤ n1 + m0 ∧ m + n2 ≤ n1 + m0 → copy(ar0, 0, ar, m, n2)[n1 + m0] = ar[n1 + m0], n1 < # ar1, n1 < n0, m0 + n0 ≤ m, m + n2 +1 ≤ # ar, n0 ≤ # ar1, n2 +1 ≤ # ar0, # ar ≤ n1 + m0, m + n2 +1 ≤ n1 + m0 ⊦ copy(ar0, 0, ar, m, n2 +1)[n1 + m0] = ar[n1 + m0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m + n2 < n1 + m0, # ar ≤ n1 + m0, n2 < # ar0, n0 ≤ # ar1, m + n2 < # ar, m0 + n0 ≤ m, n1 < n0, n1 < # ar1, ∀ n1, ar1, n0, m0, m, ar, ar0. n1 < # ar1 ∧ n1 < n0 ∧ m0 + n0 ≤ m ∧ m + n2 ≤ # ar ∧ n0 ≤ # ar1 ∧ n2 ≤ # ar0 ∧ # ar ≤ n1 + m0 ∧ m + n2 ≤ n1 + m0 → copy(ar0, 0, ar, m, n2)[n1 + m0] = ar[n1 + m0] ⊦ copy(ar0, 0, ar, m, n2)[n1 + m0] = ar[n1 + m0]

Rule argument

Additional information

Used simplifier rules: Copy-succ: ⊦ copy(ar, m, ar0, m0, n + 1) = copy(ar, m, ar0, m0, n)[m0 + n, ar[m + n]];%cngetput: m ≠ n ⊦ ar[m, a][n] = ar[n]; from specification array%ref-1: ⊦ ¬ n < n; from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ 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%add-base: ⊦ n + 0 = n; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2