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Proof step 3 from proof cons-02 (previous step)

disj(x, y) ⊦ disj(y, a ' + x) ↔ ¬ a ∈ y

User applied rule structural induction

And obtained the following new goals

Proof step 4:
disj(x, a0 + y0), ∀ x, a. disj(x, y0) → (disj(y0, a ' + x) ↔ ¬ a ∈ y0) ⊦ disj(a0 + y0, a ' + x) ↔ ¬ a ∈ a0 + y0

Rule argument

indvar: y

Additional information

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