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

ri.lebs ⊆ ns ⊦ (ri -- key).lebs ⊆ ri.lebs, ∀ n. n ∈ (ri -- key).lebs → n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
adr ∈ ri -- key, ri.lebs ⊆ ns ⊦ adr.lnum ∈ ns, (ri -- key).lebs ⊆ ri.lebs

Rule argument

Additional information

Used simplifier rules: lebs-index: ⊦ n ∈ ri.lebs ↔ (∃ adr. adr.lnum = n ∧ adr ∈ ri);