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Proof step 1 from proof firstoc (conclusion)

a ∈ x, a ∈ x0 ⊦ ∃ b, y, y0. b ∈ x ∧ x0 = y + b ' + y0 ∧ disj(x, y)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ a, x. a ∈ x ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)), a ∈ x, a ∈ a0 + x1 ⊦ ∃ b, y, y0. b ∈ x ∧ a0 + x1 = y + b ' + y0 ∧ disj(x, y)

Rule argument

indvar: x0

Additional information

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