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Proof step 2 from proof in-01 (previous step)

⊦ a ∈ b ' + x ↔ a ∈ x, a = b

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-06: ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x);