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Proof step 1 from proof app-02 (conclusion)

⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ disj(x, (a + y0) + z) ↔ disj(x, a + y0) ∧ disj(x, z)

Rule argument

term: y

Additional information

Used simplifier rules: e: ⊦ disj(x, []);%list freely generated by [], +;;%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list