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Proof step 3 from proof app (previous step)

disj(x, y + z) ⊦ disj(x, y)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
disj(x, (a + y0) + z), ∀ x, z. disj(x, y0 + z) → disj(x, y0) ⊦ disj(x, a + y0)

Rule argument

term: y

Additional information

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