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Proof step 7 from proof lebs-del-subseteq-set (previous step)

adr0 = @(m, m0, m1), adr0 ∈ as, as.lebs ⊆ ns ⊦ # as = 0, adr0 = adr, as = ∅, m ∈ ns, (as -- adr).lebs ⊆ as.lebs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information