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Proof step 1 from proof diff-in-one (conclusion)

⊦ a ∈ x \ a0 ' ↔ a ∈ x ∧ a ≠ a0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: diff-in: ⊦ a ∈ x \ y ↔ a ∈ x ∧ ¬ a ∈ y;% ⊦ a ∈ b ' ↔ a = b; (s)