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Proof step 1 from proof eq-subset (conclusion)

⊦ s = s0 ↔ s ⊆ s0 ∧ s0 ⊆ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: subset-equal: ⊦ s ⊆ s0 → (s0 ⊆ s ↔ s = s0);%sub-idem: ⊦ s ⊆ s;