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Proof step 4 from proof toSet-diff-old (previous step)

⊦ a ∈ x0, ∀ a0. a0 ∈ (s0 ++ a) \ x0.toSet ↔ a0 ∈ (s0 \ x0.toSet) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ a0 ∈ (s0 ++ a) \ x0.toSet ↔ a0 = a ∨ a0 ∈ s0 \ x0.toSet, a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)