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Proof step 4 from proof app-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z), a ∈ x

Rule argument

Additional information

Used simplifier rules: first: ⊦ ¬ a ∈ x → (disj(x, a ' + y) ↔ disj(x, y));%a: ⊦ (x + y) + z = x + y + z; from specification list