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Proof step 5 from proof range-eq-bounds (previous step)

range=(ar0, n0, ar1, n1, m0 +1), ∀ ar0, n0, ar1, n1. range=(ar0, n0, ar1, n1, m0) → n0 + m0 ≤ # ar0 ∧ n1 + m0 ≤ # ar1 ⊦ n0 + m0 +1 ≤ # ar0 ∧ n1 + m0 +1 ≤ # ar1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ range=(ar0, n0, ar1, n1, m + 1) ↔ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m)
Proof step 7:
ar0[n0] = ar1[n1], range=(ar0, n0 + 1, ar1, n1 + 1, m0), ∀ ar0, n0, ar1, n1. range=(ar0, n0, ar1, n1, m0) → n0 + m0 ≤ # ar0 ∧ n1 + m0 ≤ # ar1 ⊦ n0 + m0 < # ar0 ∧ n1 + m0 < # ar1

Rule argument

Additional information

Used simplifier rules: range-eq-rec: ⊦ range=(ar0, n0, ar1, n1, m + 1) ↔ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m);% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)