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Proof step 9 from proof in-below (previous step)

n0 ∈ ns, ∀ ns0, n, m. # ns0 < # ns → true → (n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m) ⊦ # ns = 0, n ∈ below(ns, m) ↔ n ∈ ns ∧ n < m

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ n ∈ ns ↔ (∃ ns0. ns = ns0 ++ n ∧ ¬ n ∈ ns0)
Proof step 11:
∃ ns0. ns = ns0 ++ n0 ∧ ¬ n0 ∈ ns0, ∀ ns0, n, m. # ns0 < # ns → true → (n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m) ⊦ # ns = 0, n ∈ below(ns, m) ↔ n ∈ ns ∧ n < m

Rule argument

spec: set-basic, inst: nat-set name: indef seq: ⊦ n ∈ ns ↔ (∃ ns0. ns = ns0 ++ n ∧ ¬ n ∈ ns0) subst: [ns, n] → [ns, n0] flags: <all paths>

Additional information

Used terms: ns, n0