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

as.lebs ⊆ ns ⊦ (as -- adr).lebs ⊆ as.lebs, ∀ n. n ∈ (as -- adr).lebs → n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
adr0 ∈ as, as.lebs ⊆ ns ⊦ # as = 0, adr0 = adr, as = ∅, adr0.lnum ∈ ns, (as -- adr).lebs ⊆ as.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]% ⊦ adr ∈ as -- adr0 ↔ adr ≠ adr0 ∧ adr ∈ as; (s)