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Proof step 3 from proof in-append (previous step)

∀ a, y. a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y ⊦ a ∈ (a0 + x0) + y ↔ a ∈ a0 + x0 ∨ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a, y. a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y ⊦ a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y, a = a0

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);