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Proof step 4 from proof range-eq-subarrays (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 + m ≤ # ar0, n1 + m ≤ # ar1 ⊦ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n]) ↔ subarray(ar0, n0, m) = subarray(ar1, n1, m)

Proof step 4 from proof range-eq-subarrays (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information