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Proof step 2 from proof copy-shift-append (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 4:
# ar0 ≤ n, n + m1 ≤ # ar0 + # ar1 ⊦ copy(ar0 + ar1, n, ar2, m0, m1) = copy(ar1, n - # ar0, ar2, m0, m1)
Proof step 5:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1

Rule argument

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

Additional information

Used simplifier rules: copy-append: # ar0 ≤ n, n + m1 ≤ # ar0 + # ar1 ⊦ copy(ar0 + ar1, n, ar2, m0, m1) = copy(ar1, n - # ar0, ar2, m0, m1);%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m ≤ m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m - m = 0; (s)% ⊦ # 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)% ⊦ n + 0 = n; (s) Used terms: copy(ar1, 0, ar, n, m), copy(ar0 + ar1, # ar0, ar, n, m)