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Proof step 6 from proof app-02 (previous step)

disj(x, y0), ∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z), a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information