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Proof step 3 from proof lebs-del-subseteq-set (previous step)

as.lebs ⊆ ns ⊦ (as -- adr).lebs ⊆ ns, (as -- adr).lebs ⊆ as.lebs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0)
Proof step 5:
as.lebs ⊆ ns ⊦ (as -- adr).lebs ⊆ as.lebs, ∀ n. n ∈ (as -- adr).lebs → n ∈ ns

Rule argument

spec: set-basic, inst: nat-set name: Subset seq: ⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0) subst: [ns0, ns] → [ns, (as -- adr).lebs] flags: <all paths>

Additional information

Used terms: ns, (as -- adr).lebs