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Proof step 8 from proof lebs-del-subseteq-set (previous step)

@(m, m0, m1) ∈ as, as.lebs ⊆ ns ⊦ # as = 0, @(m, m0, m1) = adr, as = ∅, m ∈ ns, (as -- adr).lebs ⊆ as.lebs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0)
Proof step 10:
@(m, m0, m1) ∈ as, ∀ n. n ∈ as.lebs → n ∈ ns ⊦ # as = 0, @(m, m0, m1) = adr, as = ∅, m ∈ ns, (as -- adr).lebs ⊆ as.lebs

Rule argument

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

Additional information

Used terms: ns, as.lebs