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Proof step 1 from proof in-01 (conclusion)

⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in-10: ⊦ ¬ a ∈ x → (a ∈ b ' + x ↔ a = b);%a: ⊦ (x + y) + z = x + y + z;%in-12: ⊦ a ∈ y → a ∈ x + y;