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Proof step 4 from proof lebs-range-eq (previous step)

range=(avol0, m, avol1, m, m2), m ≤ n0, n0 + n +1 ≤ m + m2, ∀ avol0, m, avol1, m2, n0. range=(avol0, m, avol1, m, m2) ∧ n0 + n ≤ m + m2 ∧ m ≤ n0 → lebs(avol0, n0, n) = lebs(avol1, n0, n) ⊦ lebs(avol0, n0, n +1) = lebs(avol1, n0, n +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 + n < m + m2, range=(avol0, m, avol1, m, m2), m ≤ n0, ∀ avol0, m, avol1, m2, n0. range=(avol0, m, avol1, m, m2) ∧ n0 + n ≤ m + m2 ∧ m ≤ n0 → lebs(avol0, n0, n) = lebs(avol1, n0, n) ⊦ lebs(avol0, n0, n + 1) = lebs(avol1, n0, n + 1)

Rule argument

Additional information

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