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Proof step 3 from proof inter-subset-rem-empty (previous step)

ns0 ∩ ns1 = ∅, ns2 ⊆ ns1 ⊦ #((ns0 \ ns3) ∩ ns2) = 0, (ns0 \ ns3) ∩ ns2 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0)

Rule argument

spec: set-basic, inst: nat-set name: Extension seq: ⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0) subst: [ns0, ns] → [∅, (ns0 \ ns3) ∩ ns2] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ n ∈ ∅; (s)% ⊦ n ∈ ns ∩ ns0 ↔ n ∈ ns ∧ n ∈ ns0; (ws)%int-a: ⊦ (ns ∩ ns0) ∩ ns1 = ns ∩ ns0 ∩ ns1; from specification set-union[nat-set]%int-c: ⊦ ns ∩ ns0 = ns0 ∩ ns; from specification set-union[nat-set]% ⊦ n ∈ ns \ ns0 ↔ n ∈ ns ∧ ¬ n ∈ ns0; (ws)%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]%sub-in: ⊦ n ∈ ns ∧ ns ⊆ ns0 → n ∈ ns0; (forward) from specification set-basic[nat-set]% ⊦ ns ∩ ns0 = ∅ ∧ n ∈ ns → (n ∈ ns0 ↔ false); (as) Used terms: ∅, (ns0 \ ns3) ∩ ns2