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Proof step 12 from proof elems-subseteq-add-diff-notin (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: elems-in-fwd: ⊦ n ∈ nx ∧ elems nx ⊆ ns → n ∈ ns;%elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx;