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Proof step 10 from proof firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ a1. a1 ∈ y ∨ a1 ∈ a ' ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0), a = a0 ∨ a ∈ a ' ∨ a ∈ z ⊦ 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;%a: ⊦ (x + y) + z = x + y + z;