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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ x0, a0 ∈ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ x0 \ y, a ∈ y

Rule argument

Additional information

Used simplifier rules: notin-01: ⊦ a ∈ y → ¬ a ∈ x \ y;