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Proof step 5 from proof lastoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ z), a = a0 ∨ a ∈ x0, a ∈ x0 → x0 = y + a ' + z ∧ ¬ a ∈ z ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ z

Proof step 5 from proof lastoc (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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