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Proof step 1 from proof subarray-elim-right (conclusion)

n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 3:
n + m ≤ # ar0 ⊦ # subarray(ar0 + ar1, n, m) = # subarray(ar0, n, m) ∧ (∀ n0. n0 < # subarray(ar0 + ar1, n, m) → subarray(ar0 + ar1, n, m)[n0] = subarray(ar0, n, m)[n0])

Rule argument

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

Additional information

Used terms: subarray(ar0, n, m), subarray(ar0 + ar1, n, m)