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Proof step 8 from proof copy2append0 (previous step)

# ar1 = (n2 + n1) + n0, n0 ≤ # ar1, 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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 10:
# ar1 = (n2 + n1) + n0, 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)) ∧ (∀ n. n < # copy(ar0, 0, ar1, n0, n1) → copy(ar0, 0, ar1, n0, n1)[n] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [ subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2), copy(ar0, 0, ar1, n0, n1)] flags: <all paths>

Additional information

Used terms: subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2), copy(ar0, 0, ar1, n0, n1)