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Proof step 2 from proof a (previous step)

∀ y, z. (x0 + y) + z = x0 + y + z ⊦ ((a + x0) + y) + z = (a + x0) + y + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ y, z. (x0 + y) + z = x0 + y + z ⊦ (x0 + y) + z = x0 + y + z

Rule argument

Additional information

Used simplifier rules: Cons: ⊦ (a + x) + y = a + (x + y);% ⊦ a + x = a + y ↔ x = y; (s)