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Proof step 12 from proof lebs-del-subseteq-set (previous step)

@(m, m0, m1) ∈ as, (∃ adr. adr.lnum = m ∧ adr ∈ as) → m ∈ 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

Rule argument

Additional information