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Proof step 1 from proof unrax (conclusion)

⊦ a ∈ s ∪ s0 ↔ a ∈ s ∨ a ∈ s0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: unin: ⊦ ¬ a ∈ s → (a ∈ s ∪ s0 ↔ a ∈ s0);%unin-01: ⊦ a ∈ s ∪ s0 ↔ ¬(¬ a ∈ s ∧ ¬ a ∈ s0);