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

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), n0 + n < # avol0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 + n < m + m2, # avol0 < m + m2, range=(avol0, m, avol1, m, m2), m ≤ n0, # avol0 ≤ n0 + n, ∀ 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: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)