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Proof step 36 from proof copy2append0 (previous step)

# ar1 = (n2 + n1) + n0, n = n3 + n0, n0 < # ar1, n < # ar1, n < n0 + n1, n0 ≤ n, n0 + n1 ≤ # ar1, n1 ≤ # ar0, n1 + n0 ≤ # ar1 ⊦ n2 + n1 = 0, ar0[n3] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n], n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
# ar1 = (n2 + n1) + n0, n3 < n2 + n1, n3 < # ar0, n0 < # ar1, n3 < n1, n1 + n0 ≤ # ar1, n1 ≤ # ar0, n0 + n1 ≤ # ar1 ⊦ ar0[n3] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n3 + n0]

Rule argument

Additional information

Used simplifier rules: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)