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Proof step 3 from proof copy-exact-append (previous step)

# ar0 = # ar ⊦ copy(ar0 + ar1, 0, ar, 0, # ar) = ar0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [ar0, copy(ar0 + ar1, 0, ar, 0, # ar)] flags: <all paths>

Additional information

Used simplifier rules: append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s)%⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]); (s)% ⊦ ¬ m < 0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m - 0 = m; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s) Used terms: ar0, copy(ar0 + ar1, 0, ar, 0, # ar)