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Proof step 6 from proof firstoc (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ a ∈ x0, ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y
Proof step 9:
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

Rule argument

pos: left 3

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