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Proof step 1 from proof in-06 (conclusion)

⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%a: ⊦ (x + y) + z = x + y + z;%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x);