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Proof step 5 from proof copy2append0 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 7:
# ar1 = n3 + n0, n3 = n2 + n1, n1 + n0 ≤ # ar1, n1 ≤ # ar0, n0 + n1 ≤ # ar1, n0 ≤ # ar1 ⊦ 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, n]

Additional information

Used simplifier rules: append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ 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%# ar = n ⊦ subarray(ar, 0, n) = ar; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # ar1, n0, n