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Proof step 2 from proof inter-subset-rem-empty (previous step)

ns0 ∩ ns1 = ∅, ns2 ⊆ ns1 ⊦ (ns0 \ ns3) ∩ ns2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
ns0 ∩ ns1 = ∅, ns2 ⊆ ns1 ⊦ #((ns0 \ ns3) ∩ ns2) = 0, (ns0 \ ns3) ∩ ns2 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # ns ≠ 0 → ns ≠ ∅; (forward) from specification set-basic[nat-set]%size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]