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Proof step 3 from proof subarray-emparray (previous step)

n2 ≤ n1 ⊦ subarray(fill(mkarray(n0 + n1), empty), n0, n2) = fill(mkarray(n2), empty)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 5:
n2 ≤ n1 ⊦ # subarray(fill(mkarray(n0 + n1), empty), n0, n2) = # fill(mkarray(n2), empty) ∧ (∀ n. n < # subarray(fill(mkarray(n0 + n1), empty), n0, n2) → subarray(fill(mkarray(n0 + n1), empty), n0, n2)[n] = fill(mkarray(n2), empty)[n])

Rule argument

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

Additional information

Used terms: fill(mkarray(n2), empty), subarray(fill(mkarray(n0 + n1), empty), n0, n2)