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Proof step 5 from proof lebs-lookup-subseteq (previous step)

ri.lebs ⊆ ns ⊦ ri[ks].lebs ⊆ ri.lebs, ∀ n. n ∈ ri[ks].lebs → n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
adr ∈ ri[ks], ri.lebs ⊆ ns ⊦ # ri[ks] = 0, ri[ks] = ∅, adr.lnum ∈ ns, ri[ks].lebs ⊆ ri.lebs

Rule argument

Additional information

Used simplifier rules: lebs-set: ⊦ n ∈ as.lebs ↔ (∃ adr. adr.lnum = n ∧ adr ∈ as);%size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[node-store]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[node-store]