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Proof step 2 from proof lebs-del-subseteq-set (previous step)

as.lebs ⊆ ns ⊦ (as -- adr).lebs ⊆ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
as.lebs ⊆ ns ⊦ (as -- adr).lebs ⊆ ns, (as -- adr).lebs ⊆ as.lebs

Rule argument

Additional information

Used simplifier rules: sub-ntrans-01: ⊦ ns ⊆ ns0 ∧ ¬ ns1 ⊆ ns0 → ¬ ns1 ⊆ ns; (forward) from specification set-basic[nat-set]