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Proof step 19 from proof in-below (previous step)

n < m, n ∈ below(ns0, m), n ∈ ns0, ∀ ns1, n, m. # ns1 < #(ns0 ++ n0) → true → (n ∈ below(ns1, m) ↔ n ∈ ns1 ∧ n < m) ⊦ # below(ns0, m) = 0, # ns0 = 0, below(ns0, m) = ∅, ns0 = ∅, n ∈ (n0 < m ⊃ below(ns0, m) ++ n0;below(ns0, m)), n0 ∈ ns0

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 5

Additional information

Used simplifier rules: ⊦ n ∈ ns → n ∈ ns ++ n0; (s)