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Proof step 5 from proof subarray-append-shift-right (previous step)

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

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 + subarray(ar1, 0, n0), subarray(ar0 + ar1, 0, n0 + # ar0)] flags: <all paths>

Additional information

Used terms: ar0 + subarray(ar1, 0, n0), subarray(ar0 + ar1, 0, n0 + # ar0)