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Proof step 2 from proof union-difference (previous step)

s ⊆ s0 ⊦ s ∪ (s0 \ s) = s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s ∪ (s0 \ s) ↔ a ∈ s0

Rule argument

Additional information

Used simplifier rules: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)