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Proof step 8 from proof elems-subseteq-add-diff-notin (previous step)

elems nx1 ⊆ elems nx0 ⊦ n0 ∈ elems nx0 ++ n ∧ ¬ n0 ∈ elems nx1 ↔ n0 = n ∨ n0 ∈ elems nx0 \ elems nx1, n ∈ nx0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
elems nx1 ⊆ elems nx0, n0 = n ∨ n0 ∈ elems nx0 \ elems nx1 ⊦ n ∈ nx0, (n0 = n ∨ n0 ∈ nx0) ∧ ¬ n0 ∈ nx1

Rule argument

Additional information

Used simplifier rules: elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx;%elems-in-diff: ⊦ n ∈ nx0 ∧ ¬ n ∈ nx1 → n ∈ elems nx0 \ elems nx1;% ⊦ n ∈ ns ++ n0 ↔ n = n0 ∨ n ∈ ns; (s)% ⊦ n ≠ n0 → (n ∈ ns ++ n0 ↔ n ∈ ns); (s)