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Proof step 7 from proof range-eq-ext (previous step)

range=(ar0, n0 + 1, ar1, n1 + 1, m0) ↔ (n0 + 1) + m0 ≤ # ar0 ∧ (n1 + 1) + m0 ≤ # ar1 ∧ (∀ 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 8:
range=(ar0, n0 + 1, ar1, n1 + 1, m0) ↔ n0 + m0 < # ar0 ∧ n1 + m0 < # ar1 ∧ (∀ 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])

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)