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Proof step 11 from proof range-eq-ext (previous step)

n0 + m0 < # ar0, n1 + m0 < # ar1, range=(ar0, n0 + 1, ar1, n1 + 1, m0), ∀ n. n < m0 → ar0[(n0 + 1) + n] = ar1[(n1 + 1) + n] ⊦ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m0) ↔ n0 + m0 < # ar0 ∧ n1 + m0 < # ar1 ∧ (∀ n. ¬ m0 < n → ar0[n0 + n] = ar1[n1 + n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n0 + m0 < # ar0, n1 + m0 < # ar1, range=(ar0, n0 + 1, ar1, n1 + 1, m0), ∀ n. n < m0 → ar0[(n0 + 1) + n] = ar1[(n1 + 1) + n] ⊦ ar0[n0] = ar1[n1] ↔ (∀ n. ¬ m0 < n → ar0[n0 + n] = ar1[n1 + n])

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic