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Proof step 5 from proof in-below (previous step)

∀ ns0, n, m. # ns0 < # ns → true → (n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m) ⊦ n ∈ below(ns, m) ↔ n ∈ ns ∧ n < m, ns = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ ns0, n, m. # ns0 < # ns → true → (n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m) ⊦ # ns = 0, n ∈ below(ns, m) ↔ n ∈ ns ∧ n < m, ns = ∅

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # ns ≠ 0 → ns ≠ ∅; (forward) from specification set-basic[nat-set]%size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]