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Proof step 3 from proof elems-subseteq-add-diff-notin (previous step)

elems nx1 ⊆ elems nx0 ⊦ (elems nx0 ++ n) \ elems nx1 = (elems nx0 \ elems nx1) ++ n, n ∈ nx0

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:
elems nx1 ⊆ elems nx0 ⊦ n ∈ nx0, ∀ n0. n0 ∈ (elems nx0 ++ n) \ elems nx1 ↔ n0 ∈ (elems nx0 \ elems nx1) ++ n

Rule argument

spec: set-basic, inst: nat-set name: Extension seq: ⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0) subst: [ns0, ns] → [(elems nx0 \ elems nx1) ++ n, (elems nx0 ++ n) \ elems nx1] flags: <all paths>

Additional information

Used terms: (elems nx0 \ elems nx1) ++ n, (elems nx0 ++ n) \ elems nx1