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Proof step 8 from proof firstoc (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: [], x0

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%app-08: ⊦ a ' + x = a ' + y ↔ x = y;%in-empty: ⊦ ¬ a ∈ []; Used terms: [], x0