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Proof step 10 from proof diff-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a = a0 ∨ a ∈ x0 \ y ↔ (a = a0 ∨ a ∈ x0) ∧ ¬ a ∈ y, a0 ∈ y

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list