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Proof step 6 from proof onenotin-diff (previous step)

¬ a0 ∈ a ', ∀ a. a ' \ x0 = a ' ∨ a ∈ x0 ⊦ a ' \ x0 = a ', a = a0, a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a. a ' \ x0 = a ' ∨ a ∈ x0 ⊦ a ' \ x0 = a ', a = a0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ a ∈ b ' ↔ a = b; from specification list