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Proof step 2 from proof lastoc (previous step)

a ∈ x ⊦ ∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ z), a ∈ a0 + x0 ⊦ ∃ y, z. a0 + x0 = y + a ' + z ∧ ¬ a ∈ z

Rule argument

indvar: x

Additional information

Used simplifier rules: in-empty: ⊦ ¬ a ∈ [];%list freely generated by [], +;;