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Proof step 3 from proof firstoc (previous step)

∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a ∈ a0 + x0 ⊦ ∃ y, z. a0 + x0 = y + a ' + z ∧ ¬ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y

Rule argument

Additional information

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