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Proof step 10 from proof lebs-del-subseteq-set (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
@(m, m0, m1) ∈ as, ∀ n. (∃ adr. adr.lnum = n ∧ adr ∈ as) → n ∈ ns ⊦ # as = 0, @(m, m0, m1) = adr, as = ∅, m ∈ ns, (as -- adr).lebs ⊆ as.lebs

Rule argument

Additional information

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