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Proof step 3 from proof app-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
x0 + y = x0 + z, ∀ y, z. x0 + y = x0 + z → y = z ⊦ y = z

Rule argument

Additional information

Used simplifier rules: Cons: ⊦ (a + x) + y = a + (x + y);% ⊦ a + x = a0 + x0 ↔ a = a0 ∧ x = x0; (ws)