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Proof step 5 from proof range-eq-subarrays (previous step)

n0 + m ≤ # ar0, n1 + m ≤ # ar1 ⊦ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n]) ↔ subarray(ar0, n0, m) = subarray(ar1, n1, 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:
n1 + m ≤ # ar1, n0 + m ≤ # ar0 ⊦ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n]) ↔ # subarray(ar0, n0, m) = # subarray(ar1, n1, m) ∧ (∀ n. n < # subarray(ar0, n0, m) → subarray(ar0, n0, m)[n] = subarray(ar1, n1, m)[n])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [subarray(ar1, n1, m), subarray(ar0, n0, m)] flags: <all paths>

Additional information

Used terms: subarray(ar1, n1, m), subarray(ar0, n0, m)