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Proof step 11 from proof elems-subseteq-add-diff-notin (previous step)

elems nx1 ⊆ elems nx0, n0 = n ∨ n0 ∈ elems nx0 ∧ ¬ n0 ∈ elems nx1 ⊦ n ∈ nx0, (n0 = n ∨ n0 ∈ nx0) ∧ ¬ n0 ∈ nx1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
elems nx1 ⊆ elems nx0, n0 = n ∨ n0 ∈ nx0 ∧ ¬ n0 ∈ nx1 ⊦ n ∈ nx0, (n0 = n ∨ n0 ∈ nx0) ∧ ¬ n0 ∈ nx1

Rule argument

Additional information

Used simplifier rules: elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx;