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Proof step 6 from proof in-below (previous step)

∀ 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 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0)
Proof step 8:
∀ ns0, n, m. # ns0 < # ns → true → (n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m) ⊦ # ns = 0, n ∈ below(ns, m) ↔ n ∈ ns ∧ n < m, ∀ n. n ∈ ns ↔ n ∈ ∅

Rule argument

spec: set-basic, inst: nat-set name: Extension seq: ⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0) subst: [ns0, ns] → [∅, ns] flags: <all paths>

Additional information

Used terms: ∅, ns