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Proof step 4 from proof app (previous step)

disj(x, (a + y0) + z), ∀ x, z. disj(x, y0 + z) → disj(x, y0) ⊦ disj(x, a + y0)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: app-02: ⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z);%one-01: ⊦ disj(x, a ') ↔ ¬ a ∈ x;%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)