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Proof step 1 from proof lebs-range (conclusion)

⊦ (range ri).lebs = ri.lebs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0)

Rule argument

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

Additional information

Used simplifier rules: lebs-set: ⊦ n ∈ as.lebs ↔ (∃ adr. adr.lnum = n ∧ adr ∈ as);%lebs-index: ⊦ n ∈ ri.lebs ↔ (∃ adr. adr.lnum = n ∧ adr ∈ ri);% ⊦ adr ∈ range ri ↔ adr ∈ ri; (s) Used terms: ri.lebs, (range ri).lebs