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Proof step 4 from proof diff-in (previous step)

a0 ∈ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ (a0 ' + x0) \ y ↔ (a = a0 ∨ a ∈ x0) ∧ ¬ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%a: ⊦ (x + y) + z = x + y + z; from specification list