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Proof step 3 from proof range-eq-ext (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ ar0, n0, ar1, n1. range=(ar0, n0, ar1, n1, m0) ↔ n0 + m0 ≤ # ar0 ∧ n1 + m0 ≤ # ar1 ∧ (∀ n. n < m0 → ar0[n0 + n] = ar1[n1 + n]) ⊦ range=(ar0, n0, ar1, n1, m0 + 1) ↔ n0 + m0 < # ar0 ∧ n1 + m0 < # ar1 ∧ (∀ n. ¬ m0 < n → ar0[n0 + n] = ar1[n1 + n])

Rule argument

Additional information

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