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Proof step 1 from proof elems-minus (conclusion)

⊦ elems(nx -l n) = elems nx -- n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ 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] → [elems nx -- n, elems(nx -l n)] flags: <all paths>

Additional information

Used simplifier rules: elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx;% ⊦ n ∈ ns -- n0 ↔ n ≠ n0 ∧ n ∈ ns; (s)% ⊦ n ∈ nx -l n0 ↔ n ∈ nx ∧ n ≠ n0; (ws) Used terms: elems nx -- n, elems(nx -l n)