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Proof step 9 from proof firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ a1. a1 ∈ y ∨ a1 = a ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0) ⊦ a ∈ y, ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0

Rule argument

Additional information

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