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Proof step 3 from proof elems-in-notin-ginval (previous step)

elems nx ∩ ns = ∅, n ∈ ns, n ∈ nx ⊦ # ns = 0, ns = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0)
Proof step 5:
n ∈ nx, n ∈ ns, ∀ n. n ∈ elems nx ∩ ns ↔ n ∈ ∅ ⊦ # ns = 0, ns = ∅

Rule argument

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

Additional information

Used terms: ∅, elems nx ∩ ns