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Proof step 2 from proof lastoc (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 3:
a ∈ a0 + x1, a ∈ x0, ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ ∃ b, y, y0. b ∈ a0 + x1 ∧ x0 = y + b ' + y0 ∧ disj(a0 + x1, y0)

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;% ⊦ ¬ a ∈ []; (s)