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Proof step 4 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

User applied rule cut formula

And obtained the following new goals

Proof step 5:
∀ ns0, n, m. # ns0 < # ns → true → (n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m) ⊦ n ∈ below(ns, m) ↔ n ∈ ns ∧ n < m, ns = ∅

Rule argument

fma: ns = ∅

Additional information

Used simplifier rules: below-base: ⊦ below(∅, m) = ∅;% ⊦ ¬ n ∈ ∅; (s) Used formulas: ns = ∅