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Proof step 1 from proof del-in-01 (conclusion)

⊦ a ≠ b → (¬ a ∈ s -- b ↔ ¬ a ∈ s)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);