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Proof step 4 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n0 + n1 ≤ # ar ⊦ fill(ar, a, n0, n1) = subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 6:
# ar = n2 + n0 + n1, n0 + n1 ≤ # ar ⊦ # fill(ar, a, n0, n1) = #(subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2)) ∧ (∀ n. n < # fill(ar, a, n0, n1) → fill(ar, a, n0, n1)[n] = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, 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(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2), fill(ar, a, n0, n1)] flags: <all paths>

Additional information

Used terms: subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2), fill(ar, a, n0, n1)