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Proof step 6 from proof elems-subseteq-add-diff-notin (previous step)

elems nx1 ⊆ elems nx0 ⊦ n0 ∈ (elems nx0 ++ n) \ elems nx1 ↔ n0 = n ∨ n0 ∈ elems nx0 \ elems nx1, n ∈ nx0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n ∈ ns \ ns0 ↔ n ∈ ns ∧ ¬ n ∈ ns0
Proof step 8:
elems nx1 ⊆ elems nx0 ⊦ n0 ∈ elems nx0 ++ n ∧ ¬ n0 ∈ elems nx1 ↔ n0 = n ∨ n0 ∈ elems nx0 \ elems nx1, n ∈ nx0

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, n0] flags: <all paths>

Additional information

Used terms: elems nx1, elems nx0 ++ n, n0