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Proof step 2 from proof range-eq-subarrays (previous step)

⊦ range=(ar0, n0, ar1, n1, m) ↔ n0 + m ≤ # ar0 ∧ n1 + m ≤ # ar1 ∧ subarray(ar0, n0, m) = subarray(ar1, n1, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ range=(ar0, n0, ar1, n1, m) ↔ n0 + m ≤ # ar0 ∧ n1 + m ≤ # ar1 ∧ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n])
Proof step 4:
⊦ 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)

Rule argument

spec: <toplevel> name: range-eq-ext seq: ⊦ range=(ar0, n0, ar1, n1, m) ↔ n0 + m ≤ # ar0 ∧ n1 + m ≤ # ar1 ∧ (∀ n. n < m → ar0[n0 + n] = ar1[n1 + n]) subst: [m, n1, ar1, n0, ar0] → [m, n1, ar1, n0, ar0] flags: <all paths>

Additional information

Used terms: m, n1, ar1, n0, ar0