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Proof step 5 from proof fill-append-shift (previous step)

⊦ fill(ar0 + ar1, a, n0 + # ar0, m) = ar0 + fill(ar1, a, n0, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 7:
⊦ # fill(ar0 + ar1, a, n0 + # ar0, m) = #(ar0 + fill(ar1, a, n0, m)) ∧ (∀ n. n < # fill(ar0 + ar1, a, n0 + # ar0, m) → fill(ar0 + ar1, a, n0 + # ar0, m)[n] = (ar0 + fill(ar1, a, n0, m))[n])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [ar0 + fill(ar1, a, n0, m), fill(ar0 + ar1, a, n0 + # ar0, m)] flags: <all paths>

Additional information

Used terms: ar0 + fill(ar1, a, n0, m), fill(ar0 + ar1, a, n0 + # ar0, m)