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Proof step 3 from proof copy2append0 (previous step)

n0 + n1 ≤ # ar1, n1 ≤ # ar0 ⊦ copy(ar0, 0, ar1, n0, n1) = subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, # ar1 - n0 - n1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# ar1 - n0 = n2 + n1, n1 ≤ # ar0, n0 + n1 ≤ # ar1, n1 ≤ # ar1 - n0 ⊦ copy(ar0, 0, ar1, n0, n1) = subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# ar1 - n0, n1, n]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%fle-01: ⊦ 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; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # ar1 - n0, n1, n