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Proof step 16 from proof diff-in (previous step)

a ∈ x0 \ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a = a0, a = a0, a ∈ x0, a0 ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
a ∈ x0 \ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a = a0, a ∈ x0, a0 ∈ y

Rule argument

Additional information