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Proof step 5 from proof elems-subseteq-add-diff-notin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
elems nx1 ⊆ elems nx0 ⊦ n0 ∈ (elems nx0 ++ n) \ elems nx1 ↔ n0 = n ∨ n0 ∈ elems nx0 \ elems nx1, n ∈ nx0

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ ns ++ n0 ↔ n = n0 ∨ n ∈ ns; (s)