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Proof step 3 from proof elems-in-diff (previous step)

n ∈ nx0 ⊦ n ∈ nx1, n ∈ elems nx0 \ elems nx1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n ∈ ns \ ns0 ↔ n ∈ ns ∧ ¬ n ∈ ns0

Rule argument

spec: set-union, inst: nat-set name: Difference seq: ⊦ n ∈ ns \ ns0 ↔ n ∈ ns ∧ ¬ n ∈ ns0 subst: [ns0, ns, n] → [elems nx1, elems nx0, n] flags: <all paths>

Additional information

Used simplifier rules: elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx; Used terms: elems nx1, elems nx0, n