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Proof step 2 from proof app-02 (previous step)

∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ 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 3:
∀ 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

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list